AP Chemistry Quiz: Stoichiometry
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StoichiometryQuestion 1 of 20

Copper(II) oxide reacts with carbon monoxide according to the balanced equation:   CuO(s)+CO(g)→Cu(s)+CO2(g)\;\text{CuO}(s)+\text{CO}(g)\rightarrow \text{Cu}(s)+\text{CO}_2(g). If 2.0 mol2.0\ \text{mol} of CuO\text{CuO} reacts completely with excess CO\text{CO}, how many moles of CO2\text{CO}_2 are produced?

1.0 mol1.0\ \text{mol}
2.0 mol2.0\ \text{mol}
0.50 mol0.50\ \text{mol}
4.0 mol4.0\ \text{mol}
3.0 mol3.0\ \text{mol}
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AP Chemistry Quiz

AP Chemistry Quiz: Stoichiometry

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

What this quiz covers

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

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Copper(II) oxide reacts with carbon monoxide according to the balanced equation:   CuO(s)+CO(g)→Cu(s)+CO2(g)\;\text{CuO}(s)+\text{CO}(g)\rightarrow \text{Cu}(s)+\text{CO}_2(g). If 2.0 mol2.0\ \text{mol} of CuO\text{CuO} reacts completely with excess CO\text{CO}, how many moles of CO2\text{CO}_2 are produced?

  1. 1.0 mol1.0\ \text{mol}
  2. 2.0 mol2.0\ \text{mol} (correct answer)
  3. 0.50 mol0.50\ \text{mol}
  4. 4.0 mol4.0\ \text{mol}
  5. 3.0 mol3.0\ \text{mol}

Explanation: This question assesses the skill of stoichiometry. The balanced equation CuO + CO → Cu + CO2 provides mole ratios based on the coefficients, which are all 1 in this case. These ratios indicate that 1 mole of CuO produces 1 mole of CO2. To find the moles of CO2 from 2.0 mol of CuO with excess CO, apply the 1:1 ratio directly, resulting in 2.0 mol of CO2. A tempting distractor is 4.0 mol, which might arise from mistakenly doubling the amount due to confusion with a different equation's coefficients. Always start from the balanced equation and convert to moles before applying ratios.

Question 2

Aluminum reacts with bromine according to the balanced equation:   2Al(s)+3Br2(l)→2AlBr3(s)\;2\text{Al}(s)+3\text{Br}_2(l)\rightarrow 2\text{AlBr}_3(s). If 0.300 mol0.300\ \text{mol} of Al\text{Al} reacts completely with excess Br2\text{Br}_2, how many moles of AlBr3\text{AlBr}_3 are produced?

  1. 0.200 mol0.200\ \text{mol}
  2. 0.300 mol0.300\ \text{mol} (correct answer)
  3. 0.450 mol0.450\ \text{mol}
  4. 0.600 mol0.600\ \text{mol}
  5. 0.900 mol0.900\ \text{mol}

Explanation: This question assesses the skill of stoichiometry. The balanced equation 2Al + 3Br2 → 2AlBr3 provides mole ratios, with 2 moles of Al producing 2 moles of AlBr3, or a 1:1 ratio. These ratios relate the given moles of Al to the moles of AlBr3. For 0.300 mol of Al with excess Br2, the amount is 0.300 mol of AlBr3. A tempting distractor is 0.450 mol, possibly from mistakenly using the Br2 coefficient in the ratio. Always start from the balanced equation and convert to moles before applying ratios.

Question 3

Zinc reacts with hydrochloric acid according to the balanced equation:   Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g)\;\text{Zn}(s)+2\text{HCl}(aq)\rightarrow \text{ZnCl}_2(aq)+\text{H}_2(g). If 0.500 mol0.500\ \text{mol} of HCl\text{HCl} reacts completely with excess Zn\text{Zn}, how many moles of H2\text{H}_2 are produced?

  1. 0.125 mol0.125\ \text{mol}
  2. 0.250 mol0.250\ \text{mol} (correct answer)
  3. 0.500 mol0.500\ \text{mol}
  4. 1.00 mol1.00\ \text{mol}
  5. 0.750 mol0.750\ \text{mol}

Explanation: This question assesses the skill of stoichiometry. The balanced equation Zn + 2HCl → ZnCl2 + H2 provides mole ratios, with 2 moles of HCl producing 1 mole of H2. These ratios link the given moles of HCl to the moles of H2. For 0.500 mol of HCl with excess Zn, divide by 2 to obtain 0.250 mol of H2. A tempting distractor is 0.500 mol, perhaps from assuming a 1:1 ratio without the coefficient. Always start from the balanced equation and convert to moles before applying ratios.

Question 4

Ammonia is synthesized according to the balanced equation N2(g)+3H2(g)→2NH3(g)\text{N}_2(g)+3\text{H}_2(g)\rightarrow 2\text{NH}_3(g). If 0.80 mol0.80\ \text{mol} of N2\text{N}_2 reacts completely with excess H2\text{H}_2, how many moles of NH3\text{NH}_3 are formed?

  1. 0.27 mol0.27\ \text{mol}
  2. 0.40 mol0.40\ \text{mol}
  3. 0.80 mol0.80\ \text{mol}
  4. 1.6 mol1.6\ \text{mol} (correct answer)
  5. 2.4 mol2.4\ \text{mol}

Explanation: This question tests stoichiometry. The balanced equation N₂(g) + 3H₂(g) → 2NH₃(g) provides mole ratios, such as 1 mol N₂ to 2 mol NH₃. These ratios allow us to convert the given moles of N₂ to moles of NH₃. For 0.80 mol N₂, the calculation is 0.80 mol N₂ × (2 mol NH₃ / 1 mol N₂) = 1.6 mol NH₃. A tempting distractor is 0.80 mol, which results from mistakenly using a 1:1 ratio instead of 2:1. Always start from the balanced equation and convert to moles before applying ratios.

Question 5

Aluminum reacts with chlorine gas to form aluminum chloride according to the balanced equation below:

2 Al(s)+3 Cl2(g)→2 AlCl3(s)2\,\text{Al}(s)+3\,\text{Cl}_2(g)\rightarrow 2\,\text{AlCl}_3(s)

If 0.90 mol0.90\ \text{mol} of Cl2(g)\text{Cl}_2(g) reacts with excess Al(s)\text{Al}(s), how many moles of AlCl3(s)\text{AlCl}_3(s) are produced?

  1. 0.60 mol0.60\ \text{mol} (correct answer)
  2. 1.35 mol1.35\ \text{mol}
  3. 0.90 mol0.90\ \text{mol}
  4. 0.30 mol0.30\ \text{mol}
  5. 1.80 mol1.80\ \text{mol}

Explanation: This stoichiometry problem requires finding aluminum chloride produced from chlorine gas. The balanced equation shows that 3 mol Cl23 \text{ mol } \text{Cl}_2 produces 2 mol AlCl32 \text{ mol } \text{AlCl}_3, establishing a 3:2 ratio. Starting with 0.90 mol Cl₂: (0.90 mol Cl2)×(2 mol AlCl33 mol Cl2)=0.60 mol AlCl3(0.90 \text{ mol } \text{Cl}_2) \times \left( \frac{2 \text{ mol } \text{AlCl}_3}{3 \text{ mol } \text{Cl}_2} \right) = 0.60 \text{ mol } \text{AlCl}_3. Choice C (0.90 mol0.90 \text{ mol}) incorrectly assumes equal moles of reactant and product. Always apply the stoichiometric coefficients as a ratio to convert between different substances in the reaction.

Question 6

Calcium carbonate reacts with hydrochloric acid according to the balanced equation below:

CaCO3(s)+2 HCl(aq)→CaCl2(aq)+CO2(g)+H2O(l)\mathrm{CaCO_3(s) + 2\,HCl(aq) \rightarrow CaCl_2(aq) + CO_2(g) + H_2O(l)}

If 0.30 mol0.30\ \text{mol} of CaCO3\mathrm{CaCO_3} reacts completely with excess HCl\mathrm{HCl}, how many moles of HCl\mathrm{HCl} are consumed?

  1. 0.15 mol0.15\ \text{mol}
  2. 0.60 mol0.60\ \text{mol} (correct answer)
  3. 0.90 mol0.90\ \text{mol}
  4. 0.30 mol0.30\ \text{mol}
  5. 1.2 mol1.2\ \text{mol}

Explanation: This problem applies stoichiometry to find moles of HCl consumed in a reaction with calcium carbonate. The balanced equation CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O indicates that 1 mole of CaCO₃ reacts with 2 moles of HCl, establishing a 2:1 ratio. Starting with 0.30 mol CaCO₃, we calculate: 0.30 mol CaCO₃ × (2 mol HCl/1 mol CaCO₃) = 0.60 mol HCl. A common mistake would be assuming a 1:1 ratio and answering 0.30 mol, ignoring the coefficient of 2 for HCl. Always check the coefficients in the balanced equation carefully, as they directly determine the mole ratios needed for calculations.

Question 7

Magnesium reacts with nitrogen gas to form magnesium nitride:

3 Mg(s)+N2(g)→Mg3N2(s)\mathrm{3\,Mg(s) + N_2(g) \rightarrow Mg_3N_2(s)}

If 0.90 mol0.90\ \text{mol} of Mg\mathrm{Mg} reacts completely with excess N2\mathrm{N_2}, how many moles of Mg3N2\mathrm{Mg_3N_2} are formed?

  1. 0.90 mol0.90\ \text{mol}
  2. 0.30 mol0.30\ \text{mol} (correct answer)
  3. 1.2 mol1.2\ \text{mol}
  4. 2.7 mol2.7\ \text{mol}
  5. 0.60 mol0.60\ \text{mol}

Explanation: This problem uses stoichiometry to calculate magnesium nitride formation from magnesium metal. The balanced equation 3Mg + N₂ → Mg₃N₂ reveals that 3 moles of Mg produce 1 mole of Mg₃N₂, establishing a 1:3 ratio. From 0.90 mol Mg, we calculate: 0.90 mol Mg × (1 mol Mg₃N₂/3 mol Mg) = 0.30 mol Mg₃N₂. A common error would be to multiply by 3 instead of divide, giving 2.7 mol, which reverses the relationship between reactant and product. Always verify your mole ratio by checking that units cancel properly: mol Mg × (mol Mg₃N₂/mol Mg) = mol Mg₃N₂.

Question 8

The balanced equation for the combustion of propane is shown below.

C3H8(g)+5 O2(g)→3 CO2(g)+4 H2O(g)\mathrm{C_3H_8(g) + 5\,O_2(g) \rightarrow 3\,CO_2(g) + 4\,H_2O(g)}

If 2.0 mol2.0\ \text{mol} of C3H8\mathrm{C_3H_8} reacts completely with excess O2\mathrm{O_2}, how many moles of CO2\mathrm{CO_2} are produced?

  1. 0.67 mol0.67\ \text{mol}
  2. 8.0 mol8.0\ \text{mol}
  3. 6.0 mol6.0\ \text{mol} (correct answer)
  4. 3.0 mol3.0\ \text{mol}
  5. 10 mol10\ \text{mol}

Explanation: This problem requires stoichiometry to determine the moles of CO₂ produced from propane combustion. The balanced equation C₃H₈ + 5O₂ → 3CO₂ + 4H₂O shows that 1 mole of C₃H₈ produces 3 moles of CO₂, giving a mole ratio of 3:1. To find moles of CO₂ from 2.0 mol C₃H₈, multiply: 2.0 mol C₃H₈ × (3 mol CO₂/1 mol C₃H₈) = 6.0 mol CO₂. A common error would be to use the coefficient of O₂ (5) instead of CO₂ (3), which would incorrectly give 10 mol as the answer. Always identify the correct mole ratio from the balanced equation by finding the coefficients of your given and desired substances.

Question 9

Magnesium reacts with oxygen according to the balanced equation 2Mg(s)+O2(g)→2MgO(s)\mathrm{2Mg(s) + O_2(g) \rightarrow 2MgO(s)}. If 1.00 mol1.00\ \text{mol} of Mg\mathrm{Mg} reacts with excess O2\mathrm{O_2}, how many moles of O2\mathrm{O_2} are consumed?

  1. 0.25 mol0.25\ \text{mol}
  2. 0.50 mol0.50\ \text{mol} (correct answer)
  3. 1.00 mol1.00\ \text{mol}
  4. 2.00 mol2.00\ \text{mol}
  5. 4.00 mol4.00\ \text{mol}

Explanation: This problem requires stoichiometry to find how much oxygen is consumed when magnesium reacts. The balanced equation 2Mg(s) + O₂(g) → 2MgO(s) provides the mole ratio: 2 moles of Mg react with 1 mole of O₂. Starting with 1.00 mol Mg, we calculate: 1.00 mol Mg × (1 mol O₂)/(2 mol Mg) = 0.50 mol O₂. A common error would be to use the coefficient 2 incorrectly, perhaps thinking that 1.00 mol Mg requires 2.00 mol O₂ (answer D), which reverses the actual ratio. Always write out the mole ratio as a fraction with the desired substance in the numerator and the given substance in the denominator, using coefficients from the balanced equation.

Question 10

Hydrogen peroxide decomposes according to the balanced equation 2H2O2(aq)→2H2O(l)+O2(g)\mathrm{2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)}. If 0.80 mol0.80\ \text{mol} of H2O2\mathrm{H_2O_2} decomposes completely, how many moles of O2\mathrm{O_2} are produced?

  1. 0.20 mol0.20\ \text{mol}
  2. 0.40 mol0.40\ \text{mol} (correct answer)
  3. 0.80 mol0.80\ \text{mol}
  4. 1.60 mol1.60\ \text{mol}
  5. 2.40 mol2.40\ \text{mol}

Explanation: This problem involves stoichiometry in a decomposition reaction to find moles of oxygen produced. The balanced equation 2H₂O₂(aq) → 2H₂O(l) + O₂(g) shows that 2 moles of H₂O₂ produce 1 mole of O₂. Starting with 0.80 mol H₂O₂, we calculate: 0.80 mol H₂O₂ × (1 mol O₂)/(2 mol H₂O₂) = 0.40 mol O₂. A common error would be to use a 1:1 ratio, thinking each H₂O₂ produces one O₂, giving 0.80 mol (answer C), but the balanced equation clearly shows the 2:1 ratio. Always use the coefficients from the balanced equation to establish the correct mole ratio, even when it seems counterintuitive.

Question 11

Aluminum reacts with chlorine gas to form aluminum chloride according to the balanced equation 2Al(s)+3Cl2(g)→2AlCl3(s)\mathrm{2Al(s) + 3Cl_2(g) \rightarrow 2AlCl_3(s)}. If 0.30 mol0.30\ \text{mol} of Cl2\mathrm{Cl_2} reacts with excess Al\mathrm{Al}, how many moles of AlCl3\mathrm{AlCl_3} are produced?

  1. 0.10 mol0.10\ \text{mol}
  2. 0.20 mol0.20\ \text{mol} (correct answer)
  3. 0.30 mol0.30\ \text{mol}
  4. 0.45 mol0.45\ \text{mol}
  5. 0.60 mol0.60\ \text{mol}

Explanation: This problem uses stoichiometry to calculate moles of aluminum chloride produced from chlorine gas. The balanced equation 2Al(s) + 3Cl₂(g) → 2AlCl₃(s) provides the mole ratio: 3 moles of Cl₂ produce 2 moles of AlCl₃. Starting with 0.30 mol Cl₂, we calculate: 0.30 mol Cl₂ × (2 mol AlCl₃)/(3 mol Cl₂) = 0.20 mol AlCl₃. A common error would be to multiply 0.30 by 3/2 instead of 2/3, giving 0.45 mol (answer D), which inverts the ratio and treats Cl₂ as if it were in the numerator. Always set up the mole ratio fraction carefully, with the coefficient of the desired product in the numerator and the coefficient of the given reactant in the denominator.

Question 12

Hydrogen gas reacts with chlorine gas according to the balanced equation:   H2(g)+Cl2(g)→2HCl(g)\;\text{H}_2(g)+\text{Cl}_2(g)\rightarrow 2\text{HCl}(g). If 0.150 mol0.150\ \text{mol} of H2\text{H}_2 reacts completely with excess Cl2\text{Cl}_2, how many moles of HCl\text{HCl} are formed?

  1. 0.075 mol0.075\ \text{mol}
  2. 0.150 mol0.150\ \text{mol}
  3. 0.300 mol0.300\ \text{mol} (correct answer)
  4. 0.450 mol0.450\ \text{mol}
  5. 0.600 mol0.600\ \text{mol}

Explanation: This question assesses the skill of stoichiometry. The balanced equation H2 + Cl2 → 2HCl provides mole ratios, with 1 mole of H2 producing 2 moles of HCl. These ratios link the given moles of H2 to the moles of HCl. With 0.150 mol of H2 and excess Cl2, multiply by 2 to get 0.300 mol of HCl. A tempting distractor is 0.150 mol, which could result from forgetting to apply the coefficient and using a 1:1 ratio. Always start from the balanced equation and convert to moles before applying ratios.

Question 13

Solid calcium carbonate decomposes according to the balanced equation:   CaCO3(s)→CaO(s)+CO2(g)\;\text{CaCO}_3(s)\rightarrow \text{CaO}(s)+\text{CO}_2(g). When 10.0 g10.0\ \text{g} of CaCO3\text{CaCO}_3 decomposes completely, what mass of CO2\text{CO}_2 is produced? (Molar masses: CaCO3=100.0 g mol−1\text{CaCO}_3=100.0\ \text{g mol}^{-1}, CO2=44.0 g mol−1\text{CO}_2=44.0\ \text{g mol}^{-1}.)

  1. 44 g44\ \text{g}
  2. 2.2 g2.2\ \text{g}
  3. 4.4 g4.4\ \text{g} (correct answer)
  4. 22 g22\ \text{g}
  5. 8.8 g8.8\ \text{g}

Explanation: This question assesses the skill of stoichiometry. The balanced equation CaCO3 → CaO + CO2 provides mole ratios where 1 mole of CaCO3 produces 1 mole of CO2. These ratios allow conversion from grams of CaCO3 to moles using its molar mass, then to moles of CO2, and finally to grams of CO2 using its molar mass. For 10.0 g of CaCO3 (0.100 mol), the 1:1 ratio yields 0.100 mol of CO2, or 4.4 g. A tempting distractor is 44 g, which could result from forgetting to convert grams to moles and directly using the molar mass without the ratio. Always start from the balanced equation and convert to moles before applying ratios.

Question 14

Nitrogen gas and hydrogen gas react to form ammonia according to the balanced equation:   N2(g)+3H2(g)→2NH3(g)\;\text{N}_2(g)+3\text{H}_2(g)\rightarrow 2\text{NH}_3(g). If 0.900 mol0.900\ \text{mol} of H2\text{H}_2 reacts completely with excess N2\text{N}_2, how many moles of NH3\text{NH}_3 are produced?

  1. 0.300 mol0.300\ \text{mol}
  2. 0.450 mol0.450\ \text{mol}
  3. 0.600 mol0.600\ \text{mol} (correct answer)
  4. 0.900 mol0.900\ \text{mol}
  5. 1.80 mol1.80\ \text{mol}

Explanation: This question assesses the skill of stoichiometry. The balanced equation N2 + 3H2 → 2NH3 provides mole ratios, including 3 moles of H2 to 2 moles of NH3. These ratios connect the given moles of H2 to the moles of NH3. With 0.900 mol of H2 and excess N2, multiply by 2/3 to get 0.600 mol of NH3. A tempting distractor is 0.900 mol, which might arise from assuming a 1:1 ratio without considering coefficients. Always start from the balanced equation and convert to moles before applying ratios.

Question 15

Nitrogen dioxide dimerizes to form dinitrogen tetroxide according to the balanced equation 2NO2(g)→N2O4(g)2\text{NO}_2(g)\rightarrow \text{N}_2\text{O}_4(g). If 1.2 mol1.2\ \text{mol} of NO2\text{NO}_2 reacts completely, how many moles of N2O4\text{N}_2\text{O}_4 are produced?

  1. 0.60 mol0.60\ \text{mol} (correct answer)
  2. 1.2 mol1.2\ \text{mol}
  3. 2.4 mol2.4\ \text{mol}
  4. 3.6 mol3.6\ \text{mol}
  5. 0.40 mol0.40\ \text{mol}

Explanation: This question tests stoichiometry. The balanced equation 2NO₂(g) → N₂O₄(g) provides mole ratios, such as 2 mol NO₂ to 1 mol N₂O₄. These ratios relate the given moles of NO₂ to moles of N₂O₄. For 1.2 mol NO₂, the calculation is 1.2 mol NO₂ × (1 mol N₂O₄ / 2 mol NO₂) = 0.60 mol N₂O₄. A tempting distractor is 1.2 mol, resulting from assuming a 1:1 ratio without considering the coefficient 2. Always start from the balanced equation and convert to moles before applying ratios.

Question 16

The reaction of hydrogen gas with oxygen gas forms water according to the balanced equation below:

2 H2(g)+O2(g)→2 H2O(l)2\,\text{H}_2(g)+\text{O}_2(g)\rightarrow 2\,\text{H}_2\text{O}(l)

If 3.0 mol3.0\ \text{mol} of O2(g)\text{O}_2(g) reacts with excess H2(g)\text{H}_2(g), how many moles of H2O(l)\text{H}_2\text{O}(l) are produced?

  1. 1.5 mol1.5\ \text{mol}
  2. 3.0 mol3.0\ \text{mol}
  3. 6.0 mol6.0\ \text{mol} (correct answer)
  4. 2.0 mol2.0\ \text{mol}
  5. 9.0 mol9.0\ \text{mol}

Explanation: This problem requires stoichiometry to find the moles of water produced from oxygen gas. The balanced equation shows that 1 mol O₂ produces 2 mol H₂O, giving a mole ratio of 2:1. Starting with 3.0 mol O₂, we multiply by the ratio (2 mol H₂O/1 mol O₂) to get 6.0 mol H₂O. A common mistake would be to use 3.0 mol (choice B), forgetting to apply the stoichiometric coefficient. Always start from the balanced equation and use the coefficients as mole ratios to convert between reactants and products.

Question 17

Hydrogen peroxide decomposes according to the balanced equation below:

2 H2O2(aq)→2 H2O(l)+O2(g)\mathrm{2\,H_2O_2(aq) \rightarrow 2\,H_2O(l) + O_2(g)}

If 0.80 mol0.80\ \text{mol} of H2O2\mathrm{H_2O_2} decomposes completely, how many moles of O2\mathrm{O_2} are produced?

  1. 0.80 mol0.80\ \text{mol}
  2. 0.40 mol0.40\ \text{mol} (correct answer)
  3. 1.6 mol1.6\ \text{mol}
  4. 0.20 mol0.20\ \text{mol}
  5. 2.0 mol2.0\ \text{mol}

Explanation: This question uses stoichiometry to find oxygen production from hydrogen peroxide decomposition. The balanced equation 2H₂O₂ → 2H₂O + O₂ shows that 2 moles of H₂O₂ produce 1 mole of O₂, giving a 1:2 ratio (or 0.5:1). From 0.80 mol H₂O₂, we calculate: 0.80 mol H₂O₂ × (1 mol O₂/2 mol H₂O₂) = 0.40 mol O₂. Students might mistakenly think the 2:2 ratio for H₂O₂ to H₂O means everything is 1:1, leading to the incorrect answer of 0.80 mol O₂. Always examine each specific pair of substances in the balanced equation to determine the correct mole ratio.

Question 18

Calcium carbonate decomposes according to the balanced equation CaCO3(s)→CaO(s)+CO2(g)\text{CaCO}_3(s)\rightarrow \text{CaO}(s)+\text{CO}_2(g). If 10.0 g10.0\ \text{g} of CaCO3\text{CaCO}_3 decomposes completely (molar mass CaCO3=100.0 g mol−1\text{CaCO}_3=100.0\ \text{g mol}^{-1}), how many moles of CO2\text{CO}_2 are produced?

  1. 0.050 mol0.050\ \text{mol}
  2. 0.10 mol0.10\ \text{mol} (correct answer)
  3. 0.20 mol0.20\ \text{mol}
  4. 1.0 mol1.0\ \text{mol}
  5. 10.0 mol10.0\ \text{mol}

Explanation: This question tests stoichiometry. The balanced equation CaCO₃(s) → CaO(s) + CO₂(g) provides mole ratios, such as 1 mol CaCO₃ to 1 mol CO₂. First, convert the given mass to moles: 10.0 g CaCO₃ / 100.0 g/mol = 0.10 mol CaCO₃, then use the ratio to find moles CO₂. Thus, 0.10 mol CaCO₃ × (1 mol CO₂ / 1 mol CaCO₃) = 0.10 mol CO₂. A tempting distractor is 10.0 mol, which arises from forgetting to divide by the molar mass in the unit conversion. Always start from the balanced equation and convert to moles before applying ratios.

Question 19

Iron reacts with oxygen to form iron(III) oxide:

4 Fe(s)+3 O2(g)→2 Fe2O3(s)\mathrm{4\,Fe(s) + 3\,O_2(g) \rightarrow 2\,Fe_2O_3(s)}

If 0.75 mol0.75\ \text{mol} of O2\mathrm{O_2} reacts completely with excess Fe\mathrm{Fe}, how many moles of Fe2O3\mathrm{Fe_2O_3} are produced?

  1. 0.50 mol0.50\ \text{mol} (correct answer)
  2. 1.0 mol1.0\ \text{mol}
  3. 0.25 mol0.25\ \text{mol}
  4. 1.5 mol1.5\ \text{mol}
  5. 0.38 mol0.38\ \text{mol}

Explanation: This problem uses stoichiometry to determine iron(III) oxide production from oxygen gas. The balanced equation 4Fe + 3O₂ → 2Fe₂O₃ reveals that 3 moles of O₂ produce 2 moles of Fe₂O₃, establishing a 2:3 ratio. From 0.75 mol O₂, we calculate: 0.75 mol O₂ × (2 mol Fe₂O₃/3 mol O₂) = 0.50 mol Fe₂O₃. A tempting mistake would be to use the Fe coefficient (4) somewhere in the calculation, perhaps getting 1.0 mol by incorrectly using a 4:3 ratio. Always identify which two substances you're relating (here O₂ and Fe₂O₃) and use only their coefficients to establish the mole ratio.

Question 20

The reaction of nitrogen and hydrogen to form ammonia is represented by the balanced equation N2(g)+3H2(g)→2NH3(g)\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}. If 0.60 mol0.60\ \text{mol} of H2\mathrm{H_2} reacts with excess N2\mathrm{N_2}, how many moles of NH3\mathrm{NH_3} are produced?

  1. 0.20 mol0.20\ \text{mol}
  2. 0.40 mol0.40\ \text{mol} (correct answer)
  3. 0.60 mol0.60\ \text{mol}
  4. 0.90 mol0.90\ \text{mol}
  5. 1.20 mol1.20\ \text{mol}

Explanation: This problem requires stoichiometry to determine the amount of ammonia produced from a given amount of hydrogen gas. The balanced equation N₂(g) + 3H₂(g) → 2NH₃(g) provides the mole ratio: 3 moles of H₂ produce 2 moles of NH₃. Starting with 0.60 mol H₂, we use the ratio (2 mol NH₃)/(3 mol H₂) to calculate: 0.60 mol H₂ × (2 mol NH₃)/(3 mol H₂) = 0.40 mol NH₃. A common mistake would be to use the coefficient 3 incorrectly, perhaps calculating 0.60 × 3 = 1.80 mol, which doesn't appear in the choices but shows confusion about which number goes in the numerator versus denominator. Always identify the mole ratio from the balanced equation, placing the desired substance in the numerator and the given substance in the denominator.