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Chemistry Help: Identify Limiting Reactants

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Question 1 / 10

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

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Question 1

Mg is in excess. (correct answer)
$\text{O}_2$ is in excess.
Neither is in excess; both are completely consumed.
Both are in excess because products form before reactants are used up.

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 $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ with $3$ moles $\text{H}_2$ and $1$ mole $\text{O}_2$ : if all the $\text{O}_2$ reacts ( $1$ mole), you'd need $2$ moles $\text{H}_2$ (from $2:1$ ratio), and you have $3$ moles $\text{H}_2$ —enough! But if all the $\text{H}_2$ reacts ( $3$ moles), you'd need $1.5$ moles $\text{O}_2$ (from $2:1$ ratio), and you only have $1$ mole $\text{O}_2$ —NOT enough! So $\text{O}_2$ is limiting. For $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$ with $6.0$ mol Mg and $2.0$ mol $\text{O}_2$ , dividing gives Mg: $6/2=3$ and $\text{O}_2$ : $2/1=2$ , so $\text{O}_2$ is limiting (smaller), meaning Mg is in excess. Choice A correctly identifies Mg as in excess by properly comparing the ratios showing $\text{O}_2$ runs out first. Choice C fails by assuming perfect consumption without checking the mole ratios, which reveal $\text{O}_2$ 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 $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3.$ If $2.0\ \text{mol}$ of $\text{N}_2$ and $4.0\ \text{mol}$ of $\text{H}_2$ are available, which reactant is limiting?

$\text{N}_2$ is limiting because its coefficient is 1.
$\text{H}_2$ is limiting because there is not enough to match the $1:3$ ratio. (correct answer)
$\text{N}_2$ is limiting because it has fewer moles than $\text{H}_2$ .
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 $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ with 3 moles of $\text{H}_2$ and 1 mole of $\text{O}_2$ : if all the $\text{O}_2$ reacts (1 mole), you'd need 2 moles $\text{H}_2$ (from 2:1 ratio), and you have 3 moles $\text{H}_2$ —enough! But if all the $\text{H}_2$ reacts (3 moles), you'd need 1.5 moles $\text{O}_2$ (from 2:1 ratio), and you only have 1 mole $\text{O}_2$ —NOT enough! So $\text{O}_2$ is limiting. In this case, for $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$ with 2.0 mol of $\text{N}_2$ and 4.0 mol of $\text{H}_2$ , assuming all $\text{N}_2$ reacts requires 6.0 mol $\text{H}_2$ , but only 4.0 mol is available (not enough), while assuming all $\text{H}_2$ reacts requires about 1.33 mol $\text{N}_2$ , and 2.0 mol is available (enough), so $\text{H}_2$ is limiting. Choice B correctly identifies $\text{H}_2$ 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 $\text{H}_2$ 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: $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$ with 2 moles $\text{N}_2$ , 5 moles $\text{H}_2$ . If 2 moles $\text{N}_2$ reacts (reference), needs 6 moles $\text{H}_2$ (from 1:3 ratio). Have only 5 moles $\text{H}_2$ (not enough!). $\text{H}_2$ is limiting. Alternative quick method: divide each available amount by its coefficient. Example: 2 moles $\text{N}_2$ ÷ 1 = 2. 5 moles $\text{H}_2$ ÷ 3 = 1.67. The SMALLEST result identifies limiting reactant ( $\text{H}_2$ , 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: $\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}.$ If you have $4.0\ \text{mol}$ of $\text{CH}_4$ and $6.0\ \text{mol}$ of $\text{O}_2$ , what is the maximum amount of $\text{CO}_2$ that can form (in moles)?

$2.0\ \text{mol}$
$3.0\ \text{mol}$ (correct answer)
$4.0\ \text{mol}$
$6.0\ \text{mol}$

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 $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ with 3 moles $\text{H}_2$ and 1 mole $\text{O}_2$ : if all the $\text{O}_2$ reacts (1 mole), you'd need 2 moles $\text{H}_2$ (from 2:1 ratio), and you have 3 moles $\text{H}_2$ —enough! But if all the $\text{H}_2$ reacts (3 moles), you'd need 1.5 moles $\text{O}_2$ (from 2:1 ratio), and you only have 1 mole $\text{O}_2$ —NOT enough! So $\text{O}_2$ is limiting. Here, for $\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$ with 4.0 mol $\text{CH}_4$ and 6.0 mol $\text{O}_2$ , 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 $\,\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3\,$ . If you have $2.0\ \text{mol N}_2$ and $4.0\ \text{mol H}_2$ , which reactant is limiting?

$\text{N}_2$ is limiting because it has fewer moles than $\text{H}_2$ .
$\text{H}_2$ is limiting because $2.0\ \text{mol N}_2$ would require $6.0\ \text{mol H}_2$ , but only $4.0\ \text{mol H}_2$ is available. (correct answer)
$\text{N}_2$ is limiting because the coefficient of $\text{N}_2$ is 1.
Neither is limiting because $\text{H}_2$ is in excess.

Question 5

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

$\text{Mg}$ is in excess. (correct answer)
$\text{O}_2$ is in excess.
Neither reactant is in excess; both are completely consumed.
Both reactants are in excess.

Question 6

Water forms by: $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ If $8.0\,\text{mol}$ of $\text{H}_2$ reacts with $3.0\,\text{mol}$ of $\text{O}_2$ , which statement is correct?

$\text{H}_2$ is limiting, so $\text{O}_2$ is in excess
$\text{O}_2$ is limiting, so $\text{H}_2$ is in excess (correct answer)
Both reactants are limiting
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 $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ with 3 moles $\text{H}_2$ and 1 mole $\text{O}_2$ : if all the $\text{O}_2$ reacts (1 mole), you'd need 2 moles $\text{H}_2$ (from $2:1$ ratio), and you have 3 moles $\text{H}_2$ —enough! But if all the $\text{H}_2$ reacts (3 moles), you'd need 1.5 moles $\text{O}_2$ (from $2:1$ ratio), and you only have 1 mole $\text{O}_2$ —NOT enough! So $\text{O}_2$ is limiting. With 8.0 mol $\text{H}_2$ and 3.0 mol $\text{O}_2$ , $\text{O}_2$ is limiting ( $3/1=3$ vs $8/2=4$ ), so $\text{H}_2$ is in excess. Choice B correctly states $\text{O}_2$ is limiting and $\text{H}_2$ 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: $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$ with 2 moles $\text{N}_2$ , 5 moles $\text{H}_2$ . If 2 moles $\text{N}_2$ reacts (reference), needs 6 moles $\text{H}_2$ (from $1:3$ ratio). Have only 5 moles $\text{H}_2$ (not enough!). $\text{H}_2$ is limiting. Alternative quick method: divide each available amount by its coefficient. Example: 2 moles $\text{N}_2$ ÷ 1 = 2. 5 moles $\text{H}_2$ ÷ 3 = 1.67. The SMALLEST result identifies limiting reactant ( $\text{H}_2$ , 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: $2\text{CO} + \text{O}_2 \rightarrow 2\text{CO}_2$ If $6.0\,\text{mol}$ of CO and $2.0\,\text{mol}$ of $\text{O}_2$ are available, which reactant is the limiting reactant?

CO is limiting
$\text{O}_2$ is limiting (correct answer)
Both are limiting
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 $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ with $3$ moles H $_2$ and $1$ mole O $_2$ : if all the O $_2$ reacts ( $1$ mole), you'd need $2$ moles H $_2$ (from 2:1 ratio), and you have $3$ moles H $_2$ —enough! But if all the H $_2$ reacts ( $3$ moles), you'd need $1.5$ moles O $_2$ (from 2:1 ratio), and you only have $1$ mole O $_2$ —NOT enough! So O $_2$ is limiting. Here, with $6.0$ mol CO and $2.0$ mol O $_2$ , O $_2$ is limiting because assuming all CO reacts requires $3.0$ mol O $_2$ but only $2.0$ mol is available, while assuming all O $_2$ reacts requires $4.0$ mol CO and $6.0$ mol is available (ratios 2:1). Choice B correctly identifies O $_2$ 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: $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$ 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

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

$0.60\ \text{mol NO}$
$1.0\ \text{mol NO}$
$1.2\ \text{mol NO}$ (correct answer)
$2.0\ \text{mol NO}$

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 $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ 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 ( $0.60/1=0.60$ vs $1.0/1=1$ , smallest is 0.60), then max NO is $(2 \text{ NO} / 1 \text{ O}_2) \times 0.60 = 1.2 \text{ mol}$ . 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: $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$ 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 9

Ammonia forms by the reaction $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$ If $2.0\,\text{mol}$ of $\text{N}_2$ and $4.0\,\text{mol}$ of $\text{H}_2$ are available, which reactant is limiting?

$\text{N}_2$ is limiting
$\text{H}_2$ is limiting (correct answer)
Both are limiting
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 $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ with 3 moles of $\text{H}_2$ and 1 mole of $\text{O}_2$ : if all the $\text{O}_2$ reacts (1 mole), you'd need 2 moles of $\text{H}_2$ (from $2:1$ ratio), and you have 3 moles of $\text{H}_2$ —enough! But if all the $\text{H}_2$ reacts (3 moles), you'd need 1.5 moles of $\text{O}_2$ (from $2:1$ ratio), and you only have 1 mole of $\text{O}_2$ —NOT enough! So $\text{O}_2$ is limiting. In this case, with 2.0 mol $\text{N}_2$ and 4.0 mol $\text{H}_2$ , the mole ratios show that $\text{H}_2$ is limiting because assuming all $\text{N}_2$ reacts requires 6.0 mol $\text{H}_2$ but only 4.0 mol is available, while assuming all $\text{H}_2$ reacts requires 1.33 mol $\text{N}_2$ and 2.0 mol is available. Choice B correctly identifies $\text{H}_2$ 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 $\text{N}_2$ is limiting, possibly by ignoring the 1:3 ratio and thinking fewer moles of $\text{H}_2$ mean it's excess, but the math proves $\text{H}_2$ 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: $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$ with 2 moles $\text{N}_2$ , 5 moles $\text{H}_2$ . If 2 moles $\text{N}_2$ reacts (reference), needs 6 moles $\text{H}_2$ (from $1:3$ ratio). Have only 5 moles $\text{H}_2$ (not enough!). $\text{H}_2$ is limiting. Alternative quick method: divide each available amount by its coefficient. Example: $2$ moles $\text{N}_2$ ÷ 1 = 2. $5$ moles $\text{H}_2$ ÷ 3 = 1.67. The SMALLEST result identifies limiting reactant ( $\text{H}_2$ , 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 10

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

$1.0\,\text{mol}$
$2.0\,\text{mol}$ (correct answer)
$3.0\,\text{mol}$
$6.0\,\text{mol}$

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 $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ with $3$ moles $\text{H}_2$ and $1$ mole $\text{O}_2$ : if all the $\text{O}_2$ reacts ( $1$ mole), you'd need $2$ moles $\text{H}_2$ (from $2:1$ ratio), and you have $3$ moles $\text{H}_2$ —enough! But if all the $\text{H}_2$ reacts ( $3$ moles), you'd need $1.5$ moles $\text{O}_2$ (from $2:1$ ratio), and you only have $1$ mole $\text{O}_2$ —NOT enough! So $\text{O}_2$ is limiting. In this case, with $1.0$ mol $\text{H}_2$ and $3.0$ mol $\text{Cl}_2$ , $\text{H}_2$ is limiting ( $1/1=1$ vs $3/1=3$ ), so max HCl is $2.0$ mol from the $1:2$ ratio with limiting $\text{H}_2$ . Choice B correctly gives $2.0$ mol HCl by basing the calculation on the limiting reactant $\text{H}_2$ . A distractor like Choice C might incorrectly use excess Cl2 for $3.0$ mol, but that exceeds what $\text{H}_2$ 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: $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$ with $2$ moles $\text{N}_2$ , $5$ moles $\text{H}_2$ . If $2$ moles $\text{N}_2$ reacts (reference), needs $6$ moles $\text{H}_2$ (from $1:3$ ratio). Have only $5$ moles $\text{H}_2$ (not enough!). $\text{H}_2$ is limiting. Alternative quick method: divide each available amount by its coefficient. Example: $2$ moles $\text{N}_2$ ÷ $1$ = $2$ . $5$ moles $\text{H}_2$ ÷ $3$ = $1.67$ . The SMALLEST result identifies limiting reactant ( $\text{H}_2$ , 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.