Chemistry Quiz: Identify Limiting Reactants
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Identify Limiting ReactantsQuestion 1 of 20

Magnesium oxide forms according to 2Mg+O2→2MgO2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}. If 6.0 mol6.0 \text{ mol} of Mg and 2.0 mol2.0 \text{ mol} of O2\text{O}_2 react, which reactant is in excess (left over after the reaction stops)?

Mg is in excess.
O2\text{O}_2 is in excess.
Neither is in excess; both are completely consumed.
Both are in excess because products form before reactants are used up.
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Chemistry Quiz

Chemistry Quiz: Identify Limiting Reactants

Practice Identify Limiting Reactants in Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Identify Limiting Reactants, giving you a quick way to practice the rules, question types, and explanations that matter most for Chemistry.

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Question 1

Magnesium oxide forms according to 2Mg+O2→2MgO2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}. If 6.0 mol6.0 \text{ mol} of Mg and 2.0 mol2.0 \text{ mol} of O2\text{O}_2 react, which reactant is in excess (left over after the reaction stops)?

  1. Mg is in excess. (correct answer)
  2. O2\text{O}_2 is in excess.
  3. Neither is in excess; both are completely consumed.
  4. Both are in excess because products form before reactants are used up.
Explanation: This question tests your understanding of limiting reactants—the reactant that is completely consumed first in a reaction, thereby limiting the maximum amount of product that can form. When a reaction has multiple reactants with given amounts, usually one runs out before the others—this is the limiting reactant, and it determines how much product can possibly form because once it's gone, the reaction must stop even if other reactants remain (the excess reactants). To identify limiting reactant, you must compare what you HAVE (given moles) to what you NEED (calculated from mole ratios) for each reactant: for each reactant, use the mole ratio from the balanced equation to calculate how much of it would be needed if another reactant reacted completely. Whichever reactant you don't have enough of (need more than available) is the limiting reactant! For example, in 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} with 33 moles H2\text{H}_2 and 11 mole O2\text{O}_2: if all the O2\text{O}_2 reacts (11 mole), you'd need 22 moles H2\text{H}_2 (from 2:12:1 ratio), and you have 33 moles H2\text{H}_2—enough! But if all the H2\text{H}_2 reacts (33 moles), you'd need 1.51.5 moles O2\text{O}_2 (from 2:12:1 ratio), and you only have 11 mole O2\text{O}_2—NOT enough! So O2\text{O}_2 is limiting. For 2Mg+O2→2MgO2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO} with 6.06.0 mol Mg and 2.02.0 mol O2\text{O}_2, dividing gives Mg:6/2=36/2=3 and O2\text{O}_2: 2/1=22/1=2, so O2\text{O}_2 is limiting (smaller), meaning Mg is in excess. Choice A correctly identifies Mg as in excess by properly comparing the ratios showing O2\text{O}_2 runs out first. Choice C fails by assuming perfect consumption without checking the mole ratios, which reveal O2\text{O}_2 limits the reaction. The limiting reactant identification method: (1) Write the balanced equation and identify given amounts for each reactant. (2) Pick one reactant as reference—assume all of it reacts. (3) Calculate how much of each OTHER reactant would be needed for the reference reactant to completely react (use mole ratios). (4) Compare needed vs available for each: if needed is LESS than available, that reactant is excess. If needed is MORE than available, that reactant is limiting. (5) Whichever reactant you run short on (need more than you have) is the limiting reactant! Alternative quick method: divide each available amount by its coefficient; the SMALLEST result identifies limiting reactant. Both methods work—pick whichever makes more sense to you. After identifying limiting reactant, ALWAYS use IT for product calculations, not the excess reactant! The limiting reactant determines the maximum product—keep up the great work!

Question 2

Ammonia forms according to N2+3H2→2NH3.\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3. If 2.0 mol2.0\ \text{mol} of N2\text{N}_2 and 4.0 mol4.0\ \text{mol} of H2\text{H}_2 are available, which reactant is limiting?

  1. N2\text{N}_2 is limiting because its coefficient is 1.
  2. H2\text{H}_2 is limiting because there is not enough to match the 1:31:3 ratio. (correct answer)
  3. N2\text{N}_2 is limiting because it has fewer moles than H2\text{H}_2.
  4. Both reactants are limiting and run out at the same time.
Explanation: This question tests your understanding of limiting reactants—the reactant that is completely consumed first in a reaction, thereby limiting the maximum amount of product that can form. When a reaction has multiple reactants with given amounts, usually one runs out before the others—this is the limiting reactant, and it determines how much product can possibly form because once it's gone, the reaction must stop even if other reactants remain (the excess reactants). To identify limiting reactant, you must compare what you HAVE (given moles) to what you NEED (calculated from mole ratios) for each reactant: for each reactant, use the mole ratio from the balanced equation to calculate how much of it would be needed if another reactant reacted completely. Whichever reactant you don't have enough of (need more than available) is the limiting reactant! For example, in 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} with 3 moles of H2\text{H}_2 and 1 mole of O2\text{O}_2: if all the O2\text{O}_2 reacts (1 mole), you'd need 2 moles H2\text{H}_2 (from 2:1 ratio), and you have 3 moles H2\text{H}_2—enough! But if all the H2\text{H}_2 reacts (3 moles), you'd need 1.5 moles O2\text{O}_2 (from 2:1 ratio), and you only have 1 mole O2\text{O}_2—NOT enough! So O2\text{O}_2 is limiting. In this case, for N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 with 2.0 mol of N2\text{N}_2 and 4.0 mol of H2\text{H}_2, assuming all N2\text{N}_2 reacts requires 6.0 mol H2\text{H}_2, but only 4.0 mol is available (not enough), while assuming all H2\text{H}_2 reacts requires about 1.33 mol N2\text{N}_2, and 2.0 mol is available (enough), so H2\text{H}_2 is limiting. Choice B correctly identifies H2\text{H}_2 as the limiting reactant by properly comparing needed vs available amounts using mole ratios from the balanced equation. Choice C fails by assuming fewer moles alone determine the limiting reactant, ignoring the 1:3 ratio which shows H2\text{H}_2 is insufficient. The limiting reactant identification method: (1) Write the balanced equation and identify given amounts for each reactant. (2) Pick one reactant as reference—assume all of it reacts. (3) Calculate how much of each OTHER reactant would be needed for the reference reactant to completely react (use mole ratios). (4) Compare needed vs available for each: if needed is LESS than available, that reactant is excess. If needed is MORE than available, that reactant is limiting. (5) Whichever reactant you run short on (need more than you have) is the limiting reactant! Example: N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 with 2 moles N2\text{N}_2, 5 moles H2\text{H}_2. If 2 moles N2\text{N}_2 reacts (reference), needs 6 moles H2\text{H}_2 (from 1:3 ratio). Have only 5 moles H2\text{H}_2 (not enough!). H2\text{H}_2 is limiting. Alternative quick method: divide each available amount by its coefficient. Example: 2 moles N2\text{N}_2 ÷ 1 = 2. 5 moles H2\text{H}_2 ÷ 3 = 1.67. The SMALLEST result identifies limiting reactant (H2\text{H}_2, with 1.67 < 2). This works because you're finding 'how many times can I run the reaction with each reactant?' Whichever gives the fewest runs is limiting! Both methods work—pick whichever makes more sense to you. After identifying limiting reactant, ALWAYS use IT for product calculations, not the excess reactant! The limiting reactant determines the maximum product—great job tackling this!

Question 3

Carbon dioxide forms by combustion: CH4+2O2→CO2+2H2O.\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}. If you have 4.0 mol4.0\ \text{mol} of CH4\text{CH}_4 and 6.0 mol6.0\ \text{mol} of O2\text{O}_2, what is the maximum amount of CO2\text{CO}_2 that can form (in moles)?

  1. 2.0 mol2.0\ \text{mol}
  2. 3.0 mol3.0\ \text{mol} (correct answer)
  3. 4.0 mol4.0\ \text{mol}
  4. 6.0 mol6.0\ \text{mol}
Explanation: This question tests your understanding of limiting reactants—the reactant that is completely consumed first in a reaction, thereby limiting the maximum amount of product that can form. When a reaction has multiple reactants with given amounts, usually one runs out before the others—this is the limiting reactant, and it determines how much product can possibly form because once it's gone, the reaction must stop even if other reactants remain (the excess reactants). To identify limiting reactant, you must compare what you HAVE (given moles) to what you NEED (calculated from mole ratios) for each reactant: for each reactant, use the mole ratio from the balanced equation to calculate how much of it would be needed if another reactant reacted completely. Whichever reactant you don't have enough of (need more than available) is the limiting reactant! For example, in 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} with 3 moles H2\text{H}_2 and 1 mole O2\text{O}_2: if all the O2\text{O}_2 reacts (1 mole), you'd need 2 moles H2\text{H}_2 (from 2:1 ratio), and you have 3 moles H2\text{H}_2—enough! But if all the H2\text{H}_2 reacts (3 moles), you'd need 1.5 moles O2\text{O}_2 (from 2:1 ratio), and you only have 1 mole O2\text{O}_2—NOT enough! So O2\text{O}_2 is limiting. Here, for CH4+2O2→CO2+2H2O\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} with 4.0 mol CH4\text{CH}_4 and 6.0 mol O2\text{O}_2, dividing gives CH4:4/1=4 and O2:6/2=3, so O2 is limiting (smaller value); maximum CO2 is 3.0 mol based on O2 (since 6 mol O2 produces 3 mol CO2 using the 2:1 O2:CO2 ratio). Choice B correctly identifies the maximum CO2 as 3.0 mol by properly determining O2 as limiting and calculating product from it. Choice C fails by likely using the excess reactant CH4 for calculation, which overestimates since O2 runs out first. The limiting reactant identification method: (1) Write the balanced equation and identify given amounts for each reactant. (2) Pick one reactant as reference—assume all of it reacts. (3) Calculate how much of each OTHER reactant would be needed for the reference reactant to completely react (use mole ratios). (4) Compare needed vs available for each: if needed is LESS than available, that reactant is excess. If needed is MORE than available, that reactant is limiting. (5) Whichever reactant you run short on (need more than you have) is the limiting reactant! Alternative quick method: divide each available amount by its coefficient; the SMALLEST result identifies limiting reactant. Both methods work—pick whichever makes more sense to you. After identifying limiting reactant, ALWAYS use IT for product calculations, not the excess reactant! The limiting reactant determines the maximum product—you're doing awesome!

Question 4

Ammonia forms by  N2+3H2→2NH3 \,\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3\,. If you have 2.0 mol N22.0\ \text{mol N}_2 and 4.0 mol H24.0\ \text{mol H}_2, which reactant is limiting?

  1. N2\text{N}_2 is limiting because it has fewer moles than H2\text{H}_2.
  2. H2\text{H}_2 is limiting because 2.0 mol N22.0\ \text{mol N}_2 would require 6.0 mol H26.0\ \text{mol H}_2, but only 4.0 mol H24.0\ \text{mol H}_2 is available. (correct answer)
  3. N2\text{N}_2 is limiting because the coefficient of N2\text{N}_2 is 1.
  4. Neither is limiting because H2\text{H}_2 is in excess.
Explanation: This question tests your understanding of limiting reactants—the reactant that is completely consumed first in a reaction, thereby limiting the maximum amount of product that can form. When a reaction has multiple reactants with given amounts, usually one runs out before the others—this is the limiting reactant, and it determines how much product can possibly form because once it's gone, the reaction must stop even if other reactants remain (the excess reactants). For the reaction N₂ + 3H₂ → 2NH₃ with 2.0 mol N₂ and 4.0 mol H₂: if all 2.0 mol N₂ reacts, it needs 6.0 mol H₂ (from the 1:3 ratio), but we only have 4.0 mol H₂—not enough! So H₂ is limiting. If all 4.0 mol H₂ reacts, it needs 1.33 mol N₂ (from the 3:1 ratio), and we have 2.0 mol N₂—plenty! Choice B correctly identifies H₂ as limiting by calculating that 2.0 mol N₂ would require 6.0 mol H₂, but only 4.0 mol H₂ is available. Choice A incorrectly assumes the reactant with fewer moles is always limiting without considering the stoichiometric coefficients—H₂ needs three times as many moles as N₂. Alternative quick method: divide each available amount by its coefficient. 2.0 mol N₂ ÷ 1 = 2.0. 4.0 mol H₂ ÷ 3 = 1.33. The SMALLEST result identifies limiting reactant (H₂, with 1.33 < 2.0). This works because you're finding "how many times can I run the reaction with each reactant?" Whichever gives the fewest runs is limiting!

Question 5

Magnesium oxide forms by  2Mg+O2→2MgO \,2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}\,. If you start with 6.0 mol Mg6.0\ \text{mol Mg} and 2.0 mol O22.0\ \text{mol O}_2, which reactant is in excess (left over after the reaction stops)?

  1. Mg\text{Mg} is in excess. (correct answer)
  2. O2\text{O}_2 is in excess.
  3. Neither reactant is in excess; both are completely consumed.
  4. Both reactants are in excess.
Explanation: This question tests your understanding of limiting reactants—the reactant that is completely consumed first in a reaction, thereby limiting the maximum amount of product that can form. When a reaction has multiple reactants with given amounts, usually one runs out before the others—this is the limiting reactant, and it determines how much product can possibly form because once it's gone, the reaction must stop even if other reactants remain (the excess reactants). For the reaction 2Mg + O₂ → 2MgO with 6.0 mol Mg and 2.0 mol O₂: if all 6.0 mol Mg reacts, it needs 3.0 mol O₂ (from the 2:1 ratio), but we only have 2.0 mol O₂—not enough! So O₂ is limiting. If all 2.0 mol O₂ reacts, it needs 4.0 mol Mg (from the 1:2 ratio), and we have 6.0 mol Mg—plenty! Since O₂ is limiting and Mg is needed in lesser amount than available, Mg is in excess. Choice A correctly identifies Mg as the excess reactant because when O₂ (the limiting reactant) is completely consumed, there will still be Mg left over. Choice B incorrectly identifies O₂ as excess when it's actually the limiting reactant that runs out first. The excess reactant is always the one that ISN'T limiting—after the limiting reactant is consumed and the reaction stops, the excess reactant will have some amount remaining. Quick calculation: O₂ uses 4.0 mol Mg, leaving 6.0 - 4.0 = 2.0 mol Mg in excess.

Question 6

Water forms by: 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} If 8.0 mol8.0\,\text{mol} of H2\text{H}_2 reacts with 3.0 mol3.0\,\text{mol} of O2\text{O}_2, which statement is correct?

  1. H2\text{H}_2 is limiting, so O2\text{O}_2 is in excess
  2. O2\text{O}_2 is limiting, so H2\text{H}_2 is in excess (correct answer)
  3. Both reactants are limiting
  4. Both reactants are in excess
Explanation: This question tests your understanding of limiting reactants—the reactant that is completely consumed first in a reaction, thereby limiting the maximum amount of product that can form. When a reaction has multiple reactants with given amounts, usually one runs out before the others—this is the limiting reactant, and it determines how much product can possibly form because once it's gone, the reaction must stop even if other reactants remain (the excess reactants). To identify limiting reactant, you must compare what you HAVE (given moles) to what you NEED (calculated from mole ratios) for each reactant: for each reactant, use the mole ratio from the balanced equation to calculate how much of it would be needed if another reactant reacted completely. Whichever reactant you don't have enough of (need more than available) is the limiting reactant! For example, in 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} with 3 moles H2\text{H}_2 and 1 mole O2\text{O}_2: if all the O2\text{O}_2 reacts (1 mole), you'd need 2 moles H2\text{H}_2 (from 2:12:1 ratio), and you have 3 moles H2\text{H}_2—enough! But if all the H2\text{H}_2 reacts (3 moles), you'd need 1.5 moles O2\text{O}_2 (from 2:12:1 ratio), and you only have 1 mole O2\text{O}_2—NOT enough! So O2\text{O}_2 is limiting. With 8.0 mol H2\text{H}_2 and 3.0 mol O2\text{O}_2, O2\text{O}_2 is limiting (3/1=33/1=3 vs 8/2=48/2=4), so H2\text{H}_2 is in excess. Choice B correctly states O2\text{O}_2 is limiting and H2\text{H}_2 is in excess by comparing the quotients properly. Choice A reverses it, likely from not dividing by coefficients correctly. The limiting reactant identification method: (1) Write the balanced equation and identify given amounts for each reactant. (2) Pick one reactant as reference—assume all of it reacts. (3) Calculate how much of each OTHER reactant would be needed for the reference reactant to completely react (use mole ratios). (4) Compare needed vs available for each: if needed is LESS than available, that reactant is excess. If needed is MORE than available, that reactant is limiting. (5) Whichever reactant you run short on (need more than you have) is the limiting reactant! Example: N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 with 2 moles N2\text{N}_2, 5 moles H2\text{H}_2. If 2 moles N2\text{N}_2 reacts (reference), needs 6 moles H2\text{H}_2 (from 1:31:3 ratio). Have only 5 moles H2\text{H}_2 (not enough!). H2\text{H}_2 is limiting. Alternative quick method: divide each available amount by its coefficient. Example: 2 moles N2\text{N}_2 ÷ 1 = 2. 5 moles H2\text{H}_2 ÷ 3 = 1.67. The SMALLEST result identifies limiting reactant (H2\text{H}_2, with 1.67 < 2). This works because you're finding "how many times can I run the reaction with each reactant?" Whichever gives the fewest runs is limiting! Both methods work—pick whichever makes more sense to you. After identifying limiting reactant, ALWAYS use IT for product calculations, not the excess reactant! The limiting reactant determines the maximum product.

Question 7

Carbon monoxide reacts with oxygen as: 2CO+O2→2CO22\text{CO} + \text{O}_2 \rightarrow 2\text{CO}_2 If 6.0 mol6.0\,\text{mol} of CO and 2.0 mol2.0\,\text{mol} of O2\text{O}_2 are available, which reactant is the limiting reactant?

  1. CO is limiting
  2. O2\text{O}_2 is limiting (correct answer)
  3. Both are limiting
  4. Neither is limiting
Explanation: This question tests your understanding of limiting reactants—the reactant that is completely consumed first in a reaction, thereby limiting the maximum amount of product that can form. When a reaction has multiple reactants with given amounts, usually one runs out before the others—this is the limiting reactant, and it determines how much product can possibly form because once it's gone, the reaction must stop even if other reactants remain (the excess reactants). To identify limiting reactant, you must compare what you HAVE (given moles) to what you NEED (calculated from mole ratios) for each reactant: for each reactant, use the mole ratio from the balanced equation to calculate how much of it would be needed if another reactant reacted completely. Whichever reactant you don't have enough of (need more than available) is the limiting reactant! For example, in 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} with 33 moles H2_2 and 11 mole O2_2: if all the O2_2 reacts (11 mole), you'd need 22 moles H2_2 (from 2:1 ratio), and you have 33 moles H2_2—enough! But if all the H2_2 reacts (33 moles), you'd need 1.51.5 moles O2_2 (from 2:1 ratio), and you only have 11 mole O2_2—NOT enough! So O2_2 is limiting. Here, with 6.06.0 mol CO and 2.02.0 mol O2_2, O2_2 is limiting because assuming all CO reacts requires 3.03.0 mol O2_2 but only 2.02.0 mol is available, while assuming all O2_2 reacts requires 4.04.0 mol CO and 6.06.0 mol is available (ratios 2:1). Choice B correctly identifies O2_2 as the limiting reactant through accurate ratio comparisons. Choice A fails by mistakenly claiming CO limits, possibly from not checking the needed amounts properly. The limiting reactant identification method: (1) Write the balanced equation and identify given amounts for each reactant. (2) Pick one reactant as reference—assume all of it reacts. (3) Calculate how much of each OTHER reactant would be needed for the reference reactant to completely react (use mole ratios). (4) Compare needed vs available for each: if needed is LESS than available, that reactant is excess. If needed is MORE than available, that reactant is limiting. (5) Whichever reactant you run short on (need more than you have) is the limiting reactant! Example: N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 with 2 moles N2, 5 moles H2. If 2 moles N2 reacts (reference), needs 6 moles H2 (from 1:3 ratio). Have only 5 moles H2 (not enough!). H2 is limiting. Alternative quick method: divide each available amount by its coefficient. Example: 2 moles N2 ÷ 1 = 2. 5 moles H2 ÷ 3 = 1.67. The SMALLEST result identifies limiting reactant (H2, with 1.67 < 2). This works because you're finding "how many times can I run the reaction with each reactant?" Whichever gives the fewest runs is limiting! Both methods work—pick whichever makes more sense to you. After identifying limiting reactant, ALWAYS use IT for product calculations, not the excess reactant! The limiting reactant determines the maximum product.

Question 8

Hydrogen chloride forms by the reaction  H2+Cl2→2HCl \,H_2 + Cl_2 \rightarrow 2HCl\,. If 1.0 mol H2H_2 and 3.0 mol Cl2Cl_2 are available, what is the maximum amount of HClHCl that can form?

  1. 1.0 mol HClHCl
  2. 2.0 mol HClHCl (correct answer)
  3. 3.0 mol HClHCl
  4. 6.0 mol HClHCl
Explanation: This question tests your understanding of limiting reactants—the reactant that is completely consumed first in a reaction, thereby limiting the maximum amount of product that can form. When a reaction has multiple reactants with given amounts, usually one runs out before the others—this is the limiting reactant, and it determines how much product can possibly form because once it's gone, the reaction must stop even if other reactants remain (the excess reactants). For the reaction H₂ + Cl₂ → 2HCl with 1.0 mol H₂ and 3.0 mol Cl₂: if all 1.0 mol H₂ reacts, it would need 1.0 mol Cl₂ (using the 1:1 ratio), and we have 3.0 mol Cl₂—more than enough! If all 3.0 mol Cl₂ reacts, it would need 3.0 mol H₂ (1:1 ratio), but only 1.0 mol H₂ is available—not enough! Since H₂ is limiting, we calculate the maximum HCl from it: 1.0 mol H₂ × (2 mol HCl/1 mol H₂) = 2.0 mol HCl. Choice B correctly gives 2.0 mol HCl as the maximum amount that can form when H₂ is the limiting reactant. Choice D incorrectly uses the excess reactant (Cl₂) for the calculation, which would give 6.0 mol HCl, but we can't make that much because H₂ runs out first. The limiting reactant identification method: (1) Write the balanced equation and identify given amounts for each reactant. (2) Pick one reactant as reference—assume all of it reacts. (3) Calculate how much of each OTHER reactant would be needed for the reference reactant to completely react (use mole ratios). (4) Compare needed vs available for each: if needed is MORE than available, that reactant is limiting. (5) ALWAYS use the limiting reactant to calculate the maximum product! Alternative quick method: divide each available amount by its coefficient: 1.0 mol H₂ ÷ 1 = 1.0 and 3.0 mol Cl₂ ÷ 1 = 3.0. The SMALLEST result (1.0 for H₂) identifies the limiting reactant. Since the product coefficient is 2, multiply by 2: 1.0 × 2 = 2.0 mol HCl!

Question 9

Water forms by  2H2+O2→2H2O \,2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}\,. If you have 2.0 mol H22.0\ \text{mol H}_2 and 2.0 mol O22.0\ \text{mol O}_2, how many moles of H2O\text{H}_2\text{O} can form at most?

  1. 1.0 mol H2O1.0\ \text{mol H}_2\text{O}
  2. 2.0 mol H2O2.0\ \text{mol H}_2\text{O} (correct answer)
  3. 3.0 mol H2O3.0\ \text{mol H}_2\text{O}
  4. 4.0 mol H2O4.0\ \text{mol H}_2\text{O}
Explanation: This question tests your understanding of limiting reactants—the reactant that is completely consumed first in a reaction, thereby limiting the maximum amount of product that can form. When a reaction has multiple reactants with given amounts, usually one runs out before the others—this is the limiting reactant, and it determines how much product can possibly form because once it's gone, the reaction must stop even if other reactants remain (the excess reactants). For the reaction 2H₂ + O₂ → 2H₂O with 2.0 mol H₂ and 2.0 mol O₂: if all 2.0 mol H₂ reacts, it needs 1.0 mol O₂ (from the 2:1 ratio), and we have 2.0 mol O₂—plenty! If all 2.0 mol O₂ reacts, it needs 4.0 mol H₂ (from the 1:2 ratio), but we only have 2.0 mol H₂—not enough! So H₂ is limiting. Since H₂ is limiting, we calculate product from it: 2.0 mol H₂ × (2 mol H₂O/2 mol H₂) = 2.0 mol H₂O. Choice B correctly shows 2.0 mol H₂O as the maximum amount that can form, calculated from the limiting reactant H₂. Choice D incorrectly calculates from the excess reactant O₂ (2.0 mol O₂ would give 4.0 mol H₂O if there were enough H₂, but there isn't). After identifying limiting reactant, ALWAYS use IT for product calculations, not the excess reactant! Quick verification: 2.0 mol H₂ ÷ 2 = 1.0, and 2.0 mol O₂ ÷ 1 = 2.0, so H₂ is limiting (smaller value), and it produces 2.0 mol H₂O.

Question 10

Nitrogen monoxide forms by N2+O2→2NO\text{N}_2 + \text{O}_2 \rightarrow 2\text{NO} If you start with 1.0 mol N21.0\ \text{mol N}_2 and 0.60 mol O20.60\ \text{mol O}_2, what is the maximum amount of NO\text{NO} that can form (in moles)?

  1. 0.60 mol NO0.60\ \text{mol NO}
  2. 1.0 mol NO1.0\ \text{mol NO}
  3. 1.2 mol NO1.2\ \text{mol NO} (correct answer)
  4. 2.0 mol NO2.0\ \text{mol NO}
Explanation: This question tests your understanding of limiting reactants—the reactant that is completely consumed first in a reaction, thereby limiting the maximum amount of product that can form. When a reaction has multiple reactants with given amounts, usually one runs out before the others—this is the limiting reactant, and it determines how much product can possibly form because once it's gone, the reaction must stop even if other reactants remain (the excess reactants). To identify limiting reactant, you must compare what you HAVE (given moles) to what you NEED (calculated from mole ratios) for each reactant: for each reactant, use the mole ratio from the balanced equation to calculate how much of it would be needed if another reactant reacted completely. Whichever reactant you don't have enough of (need more than available) is the limiting reactant! For example, in 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} with 3 moles H2 and 1 mole O2: if all the O2 reacts (1 mole), you'd need 2 moles H2 (from 2:1 ratio), and you have 3 moles H2—enough! But if all the H2 reacts (3 moles), you'd need 1.5 moles O2 (from 2:1 ratio), and you only have 1 mole O2—NOT enough! So O2 is limiting. For this specific reaction with 1.0 mol N2 and 0.60 mol O2, first identify O2 as limiting (0.60/1=0.600.60/1=0.60 vs 1.0/1=11.0/1=1, smallest is 0.60), then max NO is (2 NO/1 O2)×0.60=1.2 mol(2 \text{ NO} / 1 \text{ O}_2) \times 0.60 = 1.2 \text{ mol}. Choice C correctly identifies the maximum NO by using the limiting reactant O2 and the proper mole ratio. Choice B fails because it might use N2 without checking O2 shortage, but O2 limits to 1.2 mol, not 2.0. The limiting reactant identification method: (1) Write the balanced equation and identify given amounts for each reactant. (2) Pick one reactant as reference—assume all of it reacts. (3) Calculate how much of each OTHER reactant would be needed for the reference reactant to completely react (use mole ratios). (4) Compare needed vs available for each: if needed is LESS than available, that reactant is excess. If needed is MORE than available, that reactant is limiting. (5) Whichever reactant you run short on (need more than you have) is the limiting reactant! Example: N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 with 2 moles N2, 5 moles H2. If 2 moles N2 reacts (reference), needs 6 moles H2 (from 1:3 ratio). Have only 5 moles H2 (not enough!). H2 is limiting. Alternative quick method: divide each available amount by its coefficient. Example: 2 moles N2 ÷ 1 = 2. 5 moles H2 ÷ 3 = 1.67. The SMALLEST result identifies limiting reactant (H2, with 1.67 < 2). This works because you're finding "how many times can I run the reaction with each reactant?" Whichever gives the fewest runs is limiting! Both methods work—pick whichever makes more sense to you. After identifying limiting reactant, ALWAYS use IT for product calculations, not the excess reactant! The limiting reactant determines the maximum product.

Question 11

Ammonia forms by the reaction N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 If 2.0 mol2.0\,\text{mol} of N2\text{N}_2 and 4.0 mol4.0\,\text{mol} of H2\text{H}_2 are available, which reactant is limiting?

  1. N2\text{N}_2 is limiting
  2. H2\text{H}_2 is limiting (correct answer)
  3. Both are limiting
  4. Neither is limiting
Explanation: This question tests your understanding of limiting reactants—the reactant that is completely consumed first in a reaction, thereby limiting the maximum amount of product that can form. When a reaction has multiple reactants with given amounts, usually one runs out before the others—this is the limiting reactant, and it determines how much product can possibly form because once it's gone, the reaction must stop even if other reactants remain (the excess reactants). To identify limiting reactant, you must compare what you HAVE (given moles) to what you NEED (calculated from mole ratios) for each reactant: for each reactant, use the mole ratio from the balanced equation to calculate how much of it would be needed if another reactant reacted completely. Whichever reactant you don't have enough of (need more than available) is the limiting reactant! For example, in 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} with 3 moles of H2\text{H}_2 and 1 mole of O2\text{O}_2: if all the O2\text{O}_2 reacts (1 mole), you'd need 2 moles of H2\text{H}_2 (from 2:12:1 ratio), and you have 3 moles of H2\text{H}_2—enough! But if all the H2\text{H}_2 reacts (3 moles), you'd need 1.5 moles of O2\text{O}_2 (from 2:12:1 ratio), and you only have 1 mole of O2\text{O}_2—NOT enough! So O2\text{O}_2 is limiting. In this case, with 2.0 mol N2\text{N}_2 and 4.0 mol H2\text{H}_2, the mole ratios show that H2\text{H}_2 is limiting because assuming all N2\text{N}_2 reacts requires 6.0 mol H2\text{H}_2 but only 4.0 mol is available, while assuming all H2\text{H}_2 reacts requires 1.33 mol N2\text{N}_2 and 2.0 mol is available. Choice B correctly identifies H2\text{H}_2 as the limiting reactant by properly comparing needed vs available amounts using mole ratios from the balanced equation. A distractor like Choice A fails by incorrectly assuming N2\text{N}_2 is limiting, possibly by ignoring the 1:3 ratio and thinking fewer moles of H2\text{H}_2 mean it's excess, but the math proves H2\text{H}_2 limits. The limiting reactant identification method: (1) Write the balanced equation and identify given amounts for each reactant. (2) Pick one reactant as reference—assume all of it reacts. (3) Calculate how much of each OTHER reactant would be needed for the reference reactant to completely react (use mole ratios). (4) Compare needed vs available for each: if needed is LESS than available, that reactant is excess. If needed is MORE than available, that reactant is limiting. (5) Whichever reactant you run short on (need more than you have) is the limiting reactant! Example: N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 with 2 moles N2\text{N}_2, 5 moles H2\text{H}_2. If 2 moles N2\text{N}_2 reacts (reference), needs 6 moles H2\text{H}_2 (from 1:31:3 ratio). Have only 5 moles H2\text{H}_2 (not enough!). H2\text{H}_2 is limiting. Alternative quick method: divide each available amount by its coefficient. Example: 22 moles N2\text{N}_2 ÷ 1 = 2. 55 moles H2\text{H}_2 ÷ 3 = 1.67. The SMALLEST result identifies limiting reactant (H2\text{H}_2, with 1.67 < 2). This works because you're finding "how many times can I run the reaction with each reactant?" Whichever gives the fewest runs is limiting! Both methods work—pick whichever makes more sense to you. After identifying limiting reactant, ALWAYS use IT for product calculations, not the excess reactant! The limiting reactant determines the maximum product.

Question 12

Hydrogen chloride can form by: H2+Cl2→2HCl\text{H}_2 + \text{Cl}_2 \rightarrow 2\text{HCl} If you start with 1.0 mol1.0\,\text{mol} of H2\text{H}_2 and 3.0 mol3.0\,\text{mol} of Cl2\text{Cl}_2, what is the maximum amount of HCl\text{HCl} that can form (in moles)?

  1. 1.0 mol1.0\,\text{mol}
  2. 2.0 mol2.0\,\text{mol} (correct answer)
  3. 3.0 mol3.0\,\text{mol}
  4. 6.0 mol6.0\,\text{mol}
Explanation: This question tests your understanding of limiting reactants—the reactant that is completely consumed first in a reaction, thereby limiting the maximum amount of product that can form. When a reaction has multiple reactants with given amounts, usually one runs out before the others—this is the limiting reactant, and it determines how much product can possibly form because once it's gone, the reaction must stop even if other reactants remain (the excess reactants). To identify limiting reactant, you must compare what you HAVE (given moles) to what you NEED (calculated from mole ratios) for each reactant: for each reactant, use the mole ratio from the balanced equation to calculate how much of it would be needed if another reactant reacted completely. Whichever reactant you don't have enough of (need more than available) is the limiting reactant! For example, in 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} with 33 moles H2\text{H}_2 and 11 mole O2\text{O}_2: if all the O2\text{O}_2 reacts (11 mole), you'd need 22 moles H2\text{H}_2 (from 2:12:1 ratio), and you have 33 moles H2\text{H}_2—enough! But if all the H2\text{H}_2 reacts (33 moles), you'd need 1.51.5 moles O2\text{O}_2 (from 2:12:1 ratio), and you only have 11 mole O2\text{O}_2—NOT enough! So O2\text{O}_2 is limiting. In this case, with 1.01.0 mol H2\text{H}_2 and 3.03.0 mol Cl2\text{Cl}_2, H2\text{H}_2 is limiting (1/1=11/1=1 vs 3/1=33/1=3), so max HCl is 2.02.0 mol from the 1:21:2 ratio with limiting H2\text{H}_2. Choice B correctly gives 2.02.0 mol HCl by basing the calculation on the limiting reactant H2\text{H}_2. A distractor like Choice C might incorrectly use excess Cl2 for 3.03.0 mol, but that exceeds what H2\text{H}_2 allows. The limiting reactant identification method: (1) Write the balanced equation and identify given amounts for each reactant. (2) Pick one reactant as reference—assume all of it reacts. (3) Calculate how much of each OTHER reactant would be needed for the reference reactant to completely react (use mole ratios). (4) Compare needed vs available for each: if needed is LESS than available, that reactant is excess. If needed is MORE than available, that reactant is limiting. (5) Whichever reactant you run short on (need more than you have) is the limiting reactant! Example: N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 with 22 moles N2\text{N}_2, 55 moles H2\text{H}_2. If 22 moles N2\text{N}_2 reacts (reference), needs 66 moles H2\text{H}_2 (from 1:31:3 ratio). Have only 55 moles H2\text{H}_2 (not enough!). H2\text{H}_2 is limiting. Alternative quick method: divide each available amount by its coefficient. Example: 22 moles N2\text{N}_2 ÷ 11 = 22. 55 moles H2\text{H}_2 ÷ 33 = 1.671.67. The SMALLEST result identifies limiting reactant (H2\text{H}_2, with 1.67<21.67 < 2). This works because you're finding "how many times can I run the reaction with each reactant?" Whichever gives the fewest runs is limiting! Both methods work—pick whichever makes more sense to you. After identifying limiting reactant, ALWAYS use IT for product calculations, not the excess reactant! The limiting reactant determines the maximum product.

Question 13

Carbon dioxide can be produced by combustion: CH4+2O2→CO2+2H2O\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} If you have 4.0 mol4.0\,\text{mol} of CH4\text{CH}_4 and 6.0 mol6.0\,\text{mol} of O2\text{O}_2, what is the maximum amount of CO2\text{CO}_2 that can form (in moles)?

  1. 2.0 mol2.0\,\text{mol}
  2. 3.0 mol3.0\,\text{mol} (correct answer)
  3. 4.0 mol4.0\,\text{mol}
  4. 6.0 mol6.0\,\text{mol}
Explanation: This question tests your understanding of limiting reactants—the reactant that is completely consumed first in a reaction, thereby limiting the maximum amount of product that can form. When a reaction has multiple reactants with given amounts, usually one runs out before the others—this is the limiting reactant, and it determines how much product can possibly form because once it's gone, the reaction must stop even if other reactants remain (the excess reactants). To identify limiting reactant, you must compare what you HAVE (given moles) to what you NEED (calculated from mole ratios) for each reactant: for each reactant, use the mole ratio from the balanced equation to calculate how much of it would be needed if another reactant reacted completely. Whichever reactant you don't have enough of (need more than available) is the limiting reactant! For example, in 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} with 3 moles H2\text{H}_2 and 1 mole O2\text{O}_2: if all the O2\text{O}_2 reacts (1 mole), you'd need 2 moles H2\text{H}_2 (from 2:1 ratio), and you have 3 moles H2\text{H}_2—enough! But if all the H2\text{H}_2 reacts (3 moles), you'd need 1.5 moles O2\text{O}_2 (from 2:1 ratio), and you only have 1 mole O2\text{O}_2—NOT enough! So O2\text{O}_2 is limiting. Here, with 4.0 mol CH4\text{CH}_4 and 6.0 mol O2\text{O}_2, O2\text{O}_2 is limiting (6.0/2=36.0/2=3 vs 4.0/1=44.0/1=4), so max CO2\text{CO}_2 is 3.0 mol based on O2\text{O}_2 allowing 3 reactions (each producing 1 CO2\text{CO}_2). Choice B correctly identifies 3.0 mol CO2\text{CO}_2 as the maximum by using the limiting reactant O2\text{O}_2 in the calculation. A distractor like Choice C might assume CH4\text{CH}_4 limits and give 4.0 mol, but that ignores O2\text{O}_2 running out first. The limiting reactant identification method: (1) Write the balanced equation and identify given amounts for each reactant. (2) Pick one reactant as reference—assume all of it reacts. (3) Calculate how much of each OTHER reactant would be needed for the reference reactant to completely react (use mole ratios). (4) Compare needed vs available for each: if needed is LESS than available, that reactant is excess. If needed is MORE than available, that reactant is limiting. (5) Whichever reactant you run short on (need more than you have) is the limiting reactant! Example: N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 with 2 moles N2\text{N}_2, 5 moles H2\text{H}_2. If 2 moles N2\text{N}_2 reacts (reference), needs 6 moles H2\text{H}_2 (from 1:3 ratio). Have only 5 moles H2\text{H}_2 (not enough!). H2\text{H}_2 is limiting. Alternative quick method: divide each available amount by its coefficient. Example: 2 moles N2\text{N}_2 ÷ 1 = 2. 5 moles H2\text{H}_2 ÷ 3 = 1.67. The SMALLEST result identifies limiting reactant (H2\text{H}_2, with 1.67 < 2). This works because you're finding "how many times can I run the reaction with each reactant?" Whichever gives the fewest runs is limiting! Both methods work—pick whichever makes more sense to you. After identifying limiting reactant, ALWAYS use IT for product calculations, not the excess reactant! The limiting reactant determines the maximum product.

Question 14

Sulfur trioxide forms by: 2SO2+O2→2SO32\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3 If 5.0 mol5.0\,\text{mol} of SO2\text{SO}_2 and 3.0 mol3.0\,\text{mol} of O2\text{O}_2 are available, which reactant is the limiting reactant?

  1. SO2\text{SO}_2 is limiting (correct answer)
  2. O2\text{O}_2 is limiting
  3. Both are limiting
  4. Neither is limiting
Explanation: This question tests your understanding of limiting reactants—the reactant that is completely consumed first in a reaction, thereby limiting the maximum amount of product that can form. When a reaction has multiple reactants with given amounts, usually one runs out before the others—this is the limiting reactant, and it determines how much product can possibly form because once it's gone, the reaction must stop even if other reactants remain (the excess reactants). To identify limiting reactant, you must compare what you HAVE (given moles) to what you NEED (calculated from mole ratios) for each reactant: for each reactant, use the mole ratio from the balanced equation to calculate how much of it would be needed if another reactant reacted completely. Whichever reactant you don't have enough of (need more than available) is the limiting reactant! For example, in 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} with 3 moles of H2\text{H}_2 and 1 mole of O2\text{O}_2: if all the O2\text{O}_2 reacts (1 mole), you'd need 2 moles of H2\text{H}_2 (from 2:1 ratio), and you have 3 moles of H2\text{H}_2—enough! But if all the H2\text{H}_2 reacts (3 moles), you'd need 1.5 moles of O2\text{O}_2 (from 2:1 ratio), and you only have 1 mole of O2\text{O}_2—NOT enough! So O2\text{O}_2 is limiting. For 5.0 mol SO2 and 3.0 mol O2, SO2 is limiting because assuming all SO2 reacts requires 2.5 mol O2 and 3.0 mol is available, but the quotient (5/2=2.5 vs 3/1=3) shows SO2 smaller. Choice A correctly identifies SO2 as limiting by comparing the ratios accurately. Choice B might fail by assuming O2 limits due to fewer moles without dividing by coefficients. The limiting reactant identification method: (1) Write the balanced equation and identify given amounts for each reactant. (2) Pick one reactant as reference—assume all of it reacts. (3) Calculate how much of each OTHER reactant would be needed for the reference reactant to completely react (use mole ratios). (4) Compare needed vs available for each: if needed is LESS than available, that reactant is excess. If needed is MORE than available, that reactant is limiting. (5) Whichever reactant you run short on (need more than you have) is the limiting reactant! Example: N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 with 2 moles of N2\text{N}_2, 5 moles of H2\text{H}_2. If 2 moles of N2\text{N}_2 reacts (reference), needs 6 moles of H2\text{H}_2 (from 1:3 ratio). Have only 5 moles of H2\text{H}_2 (not enough!). H2\text{H}_2 is limiting. Alternative quick method: divide each available amount by its coefficient. Example: 22 moles N2\text{N}_2 ÷\div 11 = 22. 55 moles H2\text{H}_2 ÷\div 33 = 1.671.67. The SMALLEST result identifies limiting reactant (H2\text{H}_2, with 1.67<21.67 < 2). This works because you're finding "how many times can I run the reaction with each reactant?" Whichever gives the fewest runs is limiting! Both methods work—pick whichever makes more sense to you. After identifying limiting reactant, ALWAYS use IT for product calculations, not the excess reactant! The limiting reactant determines the maximum product.

Question 15

Ammonia reacts with oxygen (simplified) as: 4NH3+5O2→4NO+6H2O4\text{NH}_3 + 5\text{O}_2 \rightarrow 4\text{NO} + 6\text{H}_2\text{O} If 4.0 mol NH3\text{NH}_3 and 4.0 mol O2\text{O}_2 are mixed, which reactant is limiting?

  1. NH3\text{NH}_3 is limiting because its coefficient is 4.
  2. O2\text{O}_2 is limiting because 4.0 mol NH3\text{NH}_3 requires 5.0 mol O2\text{O}_2. (correct answer)
  3. NH3\text{NH}_3 is limiting because it has the same moles as O2\text{O}_2.
  4. Neither is limiting; equal moles means perfect ratio.
Explanation: This question tests your understanding of limiting reactants—the reactant that is completely consumed first in a reaction, thereby limiting the maximum amount of product that can form. When a reaction has multiple reactants with given amounts, usually one runs out before the others—this is the limiting reactant, and it determines how much product can possibly form because once it's gone, the reaction must stop even if other reactants remain (the excess reactants). For the reaction 4NH₃ + 5O₂ → 4NO + 6H₂O with 4.0 mol NH₃ and 4.0 mol O₂: If all 4.0 mol NH₃ reacts, it needs 5.0 mol O₂ (4.0 mol NH₃ × 5 O₂/4 NH₃ = 5.0 mol O₂), but we only have 4.0 mol O₂—NOT enough! So O₂ is limiting. If all 4.0 mol O₂ reacts, it needs 3.2 mol NH₃ (4.0 mol O₂ × 4 NH₃/5 O₂ = 3.2 mol NH₃), and we have 4.0 mol NH₃—enough! Choice B correctly identifies O₂ as limiting because 4.0 mol NH₃ would require 5.0 mol O₂ to react completely, but only 4.0 mol O₂ is available. Choice D incorrectly assumes equal moles means a perfect ratio, ignoring that the balanced equation requires a 4:5 ratio, not 1:1. The limiting reactant identification method: (1) Write the balanced equation and identify given amounts for each reactant. (2) Pick one reactant as reference—assume all of it reacts. (3) Calculate how much of each OTHER reactant would be needed for the reference reactant to completely react (use mole ratios). (4) Compare needed vs available for each: if needed is MORE than available, that reactant is limiting. Alternative quick method: divide each available amount by its coefficient: 4.0 mol NH₃ ÷ 4 = 1.0 and 4.0 mol O₂ ÷ 5 = 0.8. The SMALLEST result (0.8) identifies the limiting reactant (O₂).

Question 16

Aluminum reacts with chlorine gas: 2Al+3Cl2→2AlCl32\text{Al} + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_3 If 4.0 mol Al and 3.0 mol Cl2\text{Cl}_2 are available, what is the limiting reactant?

  1. Al is limiting because 4.0 mol Al requires 6.0 mol Cl2\text{Cl}_2.
  2. Cl2\text{Cl}_2 is limiting because 3.0 mol Cl2\text{Cl}_2 requires only 1.0 mol Al.
  3. Cl2\text{Cl}_2 is limiting because 3.0 mol Cl2\text{Cl}_2 requires 2.0 mol Al, and Al is in excess. (correct answer)
  4. Neither is limiting because both have whole-number moles.
Explanation: This question tests your understanding of limiting reactants—the reactant that is completely consumed first in a reaction, thereby limiting the maximum amount of product that can form. When a reaction has multiple reactants with given amounts, usually one runs out before the others—this is the limiting reactant, and it determines how much product can possibly form because once it's gone, the reaction must stop even if other reactants remain (the excess reactants). For the reaction 2Al + 3Cl₂ → 2AlCl₃ with 4.0 mol Al and 3.0 mol Cl₂: If all 4.0 mol Al reacts, it needs 6.0 mol Cl₂ (4.0 mol Al × 3 Cl₂/2 Al = 6.0 mol Cl₂), but we only have 3.0 mol Cl₂—NOT enough! So Cl₂ is limiting. If all 3.0 mol Cl₂ reacts, it needs 2.0 mol Al (3.0 mol Cl₂ × 2 Al/3 Cl₂ = 2.0 mol Al), and we have 4.0 mol Al—enough! So Al is in excess. Choice C correctly identifies Cl₂ as limiting because 3.0 mol Cl₂ requires only 2.0 mol Al to react completely, leaving 2.0 mol Al in excess. Choice A incorrectly identifies Al as limiting and miscalculates the required amount of Cl₂. The limiting reactant identification method: (1) Write the balanced equation and identify given amounts for each reactant. (2) Pick one reactant as reference—assume all of it reacts. (3) Calculate how much of each OTHER reactant would be needed for the reference reactant to completely react (use mole ratios). (4) Compare needed vs available for each: if needed is MORE than available, that reactant is limiting. Alternative quick method: divide each available amount by its coefficient: 4.0 mol Al ÷ 2 = 2.0 and 3.0 mol Cl₂ ÷ 3 = 1.0. The SMALLEST result (1.0) identifies the limiting reactant (Cl₂). This works because you're finding "how many times can I run the reaction with each reactant?" Whichever gives the fewest runs is limiting!

Question 17

Calcium reacts with hydrochloric acid: Ca+2HCl→CaCl2+H2\text{Ca} + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2 If you have 3.0 mol Ca and 4.0 mol HCl, what is the maximum amount of H2\text{H}_2 that can form?

  1. 1.0 mol H2\text{H}_2
  2. 2.0 mol H2\text{H}_2 (correct answer)
  3. 3.0 mol H2\text{H}_2
  4. 4.0 mol H2\text{H}_2
Explanation: This question tests your understanding of limiting reactants—the reactant that is completely consumed first in a reaction, thereby limiting the maximum amount of product that can form. When a reaction has multiple reactants with given amounts, usually one runs out before the others—this is the limiting reactant, and it determines how much product can possibly form because once it's gone, the reaction must stop even if other reactants remain (the excess reactants). For the reaction Ca + 2HCl → CaCl₂ + H₂ with 3.0 mol Ca and 4.0 mol HCl: If all 3.0 mol Ca reacts, it needs 6.0 mol HCl (3.0 mol Ca × 2 HCl/1 Ca = 6.0 mol HCl), but we only have 4.0 mol HCl—NOT enough! So HCl is limiting. If all 4.0 mol HCl reacts, it needs 2.0 mol Ca (4.0 mol HCl × 1 Ca/2 HCl = 2.0 mol Ca), and we have 3.0 mol Ca—enough! Since HCl is limiting, we calculate H₂ produced from it: 4.0 mol HCl × 1 H₂/2 HCl = 2.0 mol H₂. Choice B correctly shows 2.0 mol H₂ as the maximum amount that can form when HCl is the limiting reactant. Choice C incorrectly calculates from Ca (the excess reactant), which would require 6.0 mol HCl that we don't have. The limiting reactant identification method: (1) Write the balanced equation and identify given amounts for each reactant. (2) Pick one reactant as reference—assume all of it reacts. (3) Calculate how much of each OTHER reactant would be needed for the reference reactant to completely react (use mole ratios). (4) Compare needed vs available for each: if needed is MORE than available, that reactant is limiting. (5) ALWAYS use the limiting reactant to calculate product amounts! Alternative quick method: divide each available amount by its coefficient: 3.0 mol Ca ÷ 1 = 3.0 and 4.0 mol HCl ÷ 2 = 2.0. The SMALLEST result (2.0) identifies the limiting reactant (HCl) and tells you the reaction can occur 2.0 times, producing 2.0 mol H₂.

Question 18

Methane burns according to: CH4+2O2→CO2+2H2O\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} If 4.0 mol CH4\text{CH}_4 and 6.0 mol O2\text{O}_2 are available, what is the maximum amount of CO2\text{CO}_2 that can form?

  1. 2.0 mol CO2\text{CO}_2
  2. 3.0 mol CO2\text{CO}_2 (correct answer)
  3. 4.0 mol CO2\text{CO}_2
  4. 6.0 mol CO2\text{CO}_2
Explanation: This question tests your understanding of limiting reactants—the reactant that is completely consumed first in a reaction, thereby limiting the maximum amount of product that can form. When a reaction has multiple reactants with given amounts, usually one runs out before the others—this is the limiting reactant, and it determines how much product can possibly form because once it's gone, the reaction must stop even if other reactants remain (the excess reactants). For the reaction CH₄ + 2O₂ → CO₂ + 2H₂O with 4.0 mol CH₄ and 6.0 mol O₂: If all 4.0 mol CH₄ reacts, it needs 8.0 mol O₂ (4.0 mol CH₄ × 2 O₂/1 CH₄ = 8.0 mol O₂), but we only have 6.0 mol O₂—NOT enough! So O₂ is limiting. If all 6.0 mol O₂ reacts, it needs 3.0 mol CH₄ (6.0 mol O₂ × 1 CH₄/2 O₂ = 3.0 mol CH₄), and we have 4.0 mol CH₄—enough! Since O₂ is limiting, we calculate the CO₂ produced from it: 6.0 mol O₂ × 1 CO₂/2 O₂ = 3.0 mol CO₂. Choice B correctly shows 3.0 mol CO₂ as the maximum amount that can form when O₂ is the limiting reactant. Choice C incorrectly calculates from CH₄ (the excess reactant), and choices A and D don't match any correct stoichiometric calculation. The limiting reactant identification method: (1) Write the balanced equation and identify given amounts for each reactant. (2) Pick one reactant as reference—assume all of it reacts. (3) Calculate how much of each OTHER reactant would be needed for the reference reactant to completely react (use mole ratios). (4) Compare needed vs available for each: if needed is MORE than available, that reactant is limiting. (5) ALWAYS use the limiting reactant to calculate product amounts! Alternative quick method: divide each available amount by its coefficient: 4.0 mol CH₄ ÷ 1 = 4.0 and 6.0 mol O₂ ÷ 2 = 3.0. The SMALLEST result (3.0) identifies the limiting reactant (O₂) and tells you the maximum times the reaction can occur, producing 3.0 mol CO₂.

Question 19

Iron(III) oxide forms by the reaction  4Fe+3O2→2Fe2O3 \,4Fe + 3O_2 \rightarrow 2Fe_2O_3\,. If you have 8.0 mol FeFe and 5.0 mol O2O_2, which reactant is the limiting reactant?

  1. O2O_2 is limiting because 8.0 mol FeFe requires 6.0 mol O2O_2 (using the 4:3 ratio), but only 5.0 mol O2O_2 is available. (correct answer)
  2. FeFe is limiting because it has fewer moles than O2O_2.
  3. FeFe is limiting because 5.0 mol O2O_2 requires 10.0 mol FeFe.
  4. Neither reactant is limiting because both amounts are greater than the coefficients.
Explanation: This question tests your understanding of limiting reactants—the reactant that is completely consumed first in a reaction, thereby limiting the maximum amount of product that can form. When a reaction has multiple reactants with given amounts, usually one runs out before the others—this is the limiting reactant, and it determines how much product can possibly form because once it's gone, the reaction must stop even if other reactants remain (the excess reactants). For the reaction 4Fe + 3O₂ → 2Fe₂O₃ with 8.0 mol Fe and 5.0 mol O₂: if all 8.0 mol Fe reacts, it would need 6.0 mol O₂ (using the 4:3 ratio: 8.0 × 3/4 = 6.0), but only 5.0 mol O₂ is available—not enough! If all 5.0 mol O₂ reacts, it would need 6.67 mol Fe (5.0 × 4/3 = 6.67), and we have 8.0 mol Fe—more than enough! Since we don't have enough O₂ to react with all the Fe, O₂ is the limiting reactant. Choice A correctly identifies O₂ as limiting by properly calculating that 8.0 mol Fe would require 6.0 mol O₂, which exceeds the available 5.0 mol. Choice B incorrectly assumes Fe is limiting because it has fewer moles than O₂, but that's not even true (8.0 > 5.0), and even if it were, you must consider the stoichiometric ratios, not just the absolute amounts. The limiting reactant identification method: (1) Write the balanced equation and identify given amounts for each reactant. (2) Pick one reactant as reference—assume all of it reacts. (3) Calculate how much of each OTHER reactant would be needed for the reference reactant to completely react (use mole ratios). (4) Compare needed vs available for each: if needed is MORE than available, that reactant is limiting. Alternative quick method: divide each available amount by its coefficient: 8.0 mol Fe ÷ 4 = 2.0 and 5.0 mol O₂ ÷ 3 = 1.67. The SMALLEST result (1.67 for O₂) identifies the limiting reactant!

Question 20

Nitrogen monoxide forms by the reaction  N2+O2→2NO \,N_2 + O_2 \rightarrow 2NO\,. If 2.0 mol N2N_2 and 1.0 mol O2O_2 are available, which reactant will be completely consumed first (the limiting reactant)?

  1. N2N_2 is limiting because it has more moles to react.
  2. O2O_2 is limiting because it has fewer moles and the ratio is 1:1. (correct answer)
  3. N2N_2 is limiting because the coefficient of N2N_2 is 2.
  4. Neither is limiting; both will be used up exactly.
Explanation: This question tests your understanding of limiting reactants—the reactant that is completely consumed first in a reaction, thereby limiting the maximum amount of product that can form. When a reaction has multiple reactants with given amounts, usually one runs out before the others—this is the limiting reactant, and it determines how much product can possibly form because once it's gone, the reaction must stop even if other reactants remain (the excess reactants). For the reaction N₂ + O₂ → 2NO with 2.0 mol N₂ and 1.0 mol O₂: the balanced equation shows a 1:1 ratio between N₂ and O₂. If all 1.0 mol O₂ reacts, it would need 1.0 mol N₂ (1:1 ratio), and we have 2.0 mol N₂—more than enough! If all 2.0 mol N₂ reacts, it would need 2.0 mol O₂ (1:1 ratio), but only 1.0 mol O₂ is available—not enough! Since O₂ runs out first, it's the limiting reactant. Choice B correctly identifies O₂ as limiting because with a 1:1 ratio and O₂ having fewer moles (1.0 < 2.0), it will be consumed first. Choice C incorrectly focuses on the coefficient of N₂ being 2, but that's the coefficient for the product NO, not N₂—both reactants have coefficients of 1 in this balanced equation. The limiting reactant identification method: (1) Write the balanced equation and identify given amounts for each reactant. (2) Pick one reactant as reference—assume all of it reacts. (3) Calculate how much of each OTHER reactant would be needed for the reference reactant to completely react (use mole ratios). (4) Compare needed vs available for each: if needed is MORE than available, that reactant is limiting. Alternative quick method: divide each available amount by its coefficient: 2.0 mol N₂ ÷ 1 = 2.0 and 1.0 mol O₂ ÷ 1 = 1.0. The SMALLEST result (1.0 for O₂) identifies the limiting reactant! When coefficients are equal (both 1 here), the reactant with fewer moles is limiting.