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AP Chemistry

AP Chemistry Help: Stoichiometry

Review real example questions for Stoichiometry in AP Chemistry.

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)CuO(s)+CO(g)→Cu(s)+CO2​(g). If 2.0 mol2.0\ \text{mol}2.0 mol of CuO\text{CuO}CuO reacts completely with excess CO\text{CO}CO, how many moles of CO2\text{CO}_2CO2​ are produced?

  1. 1.0 mol1.0\ \text{mol}1.0 mol
  2. 2.0 mol2.0\ \text{mol}2.0 mol
  3. 0.50 mol0.50\ \text{mol}0.50 mol
  4. 4.0 mol4.0\ \text{mol}4.0 mol
  5. 3.0 mol3.0\ \text{mol}3.0 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)2Al(s)+3Br2​(l)→2AlBr3​(s). If 0.300 mol0.300\ \text{mol}0.300 mol of Al\text{Al}Al reacts completely with excess Br2\text{Br}_2Br2​, how many moles of AlBr3\text{AlBr}_3AlBr3​ are produced?

  1. 0.200 mol0.200\ \text{mol}0.200 mol
  2. 0.300 mol0.300\ \text{mol}0.300 mol
  3. 0.450 mol0.450\ \text{mol}0.450 mol
  4. 0.600 mol0.600\ \text{mol}0.600 mol
  5. 0.900 mol0.900\ \text{mol}0.900 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)Zn(s)+2HCl(aq)→ZnCl2​(aq)+H2​(g). If 0.500 mol0.500\ \text{mol}0.500 mol of HCl\text{HCl}HCl reacts completely with excess Zn\text{Zn}Zn, how many moles of H2\text{H}_2H2​ are produced?

  1. 0.125 mol0.125\ \text{mol}0.125 mol
  2. 0.250 mol0.250\ \text{mol}0.250 mol
  3. 0.500 mol0.500\ \text{mol}0.500 mol
  4. 1.00 mol1.00\ \text{mol}1.00 mol
  5. 0.750 mol0.750\ \text{mol}0.750 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)N2​(g)+3H2​(g)→2NH3​(g). If 0.80 mol0.80\ \text{mol}0.80 mol of N2\text{N}_2N2​ reacts completely with excess H2\text{H}_2H2​, how many moles of NH3\text{NH}_3NH3​ are formed?

  1. 0.27 mol0.27\ \text{mol}0.27 mol
  2. 0.40 mol0.40\ \text{mol}0.40 mol
  3. 0.80 mol0.80\ \text{mol}0.80 mol
  4. 1.6 mol1.6\ \text{mol}1.6 mol
  5. 2.4 mol2.4\ \text{mol}2.4 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)2Al(s)+3Cl2​(g)→2AlCl3​(s)

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

  1. 0.60 mol0.60\ \text{mol}0.60 mol
  2. 1.35 mol1.35\ \text{mol}1.35 mol
  3. 0.90 mol0.90\ \text{mol}0.90 mol
  4. 0.30 mol0.30\ \text{mol}0.30 mol
  5. 1.80 mol1.80\ \text{mol}1.80 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}_23 mol Cl2​ produces 2 mol AlCl32 \text{ mol } \text{AlCl}_32 mol AlCl3​, 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(0.90 mol Cl2​)×(3 mol Cl2​2 mol AlCl3​​)=0.60 mol AlCl3​. Choice C (0.90 mol0.90 \text{ mol}0.90 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)}CaCO3​(s)+2HCl(aq)→CaCl2​(aq)+CO2​(g)+H2​O(l)

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

  1. 0.15 mol0.15\ \text{mol}0.15 mol
  2. 0.60 mol0.60\ \text{mol}0.60 mol
  3. 0.90 mol0.90\ \text{mol}0.90 mol
  4. 0.30 mol0.30\ \text{mol}0.30 mol
  5. 1.2 mol1.2\ \text{mol}1.2 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)}3Mg(s)+N2​(g)→Mg3​N2​(s)

If 0.90 mol0.90\ \text{mol}0.90 mol of Mg\mathrm{Mg}Mg reacts completely with excess N2\mathrm{N_2}N2​, how many moles of Mg3N2\mathrm{Mg_3N_2}Mg3​N2​ are formed?

  1. 0.90 mol0.90\ \text{mol}0.90 mol
  2. 0.30 mol0.30\ \text{mol}0.30 mol
  3. 1.2 mol1.2\ \text{mol}1.2 mol
  4. 2.7 mol2.7\ \text{mol}2.7 mol
  5. 0.60 mol0.60\ \text{mol}0.60 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)}C3​H8​(g)+5O2​(g)→3CO2​(g)+4H2​O(g)

If 2.0 mol2.0\ \text{mol}2.0 mol of C3H8\mathrm{C_3H_8}C3​H8​ reacts completely with excess O2\mathrm{O_2}O2​, how many moles of CO2\mathrm{CO_2}CO2​ are produced?

  1. 0.67 mol0.67\ \text{mol}0.67 mol
  2. 8.0 mol8.0\ \text{mol}8.0 mol
  3. 6.0 mol6.0\ \text{mol}6.0 mol
  4. 3.0 mol3.0\ \text{mol}3.0 mol
  5. 10 mol10\ \text{mol}10 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)}2Mg(s)+O2​(g)→2MgO(s). If 1.00 mol1.00\ \text{mol}1.00 mol of Mg\mathrm{Mg}Mg reacts with excess O2\mathrm{O_2}O2​, how many moles of O2\mathrm{O_2}O2​ are consumed?

  1. 0.25 mol0.25\ \text{mol}0.25 mol
  2. 0.50 mol0.50\ \text{mol}0.50 mol
  3. 1.00 mol1.00\ \text{mol}1.00 mol
  4. 2.00 mol2.00\ \text{mol}2.00 mol
  5. 4.00 mol4.00\ \text{mol}4.00 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)}2H2​O2​(aq)→2H2​O(l)+O2​(g). If 0.80 mol0.80\ \text{mol}0.80 mol of H2O2\mathrm{H_2O_2}H2​O2​ decomposes completely, how many moles of O2\mathrm{O_2}O2​ are produced?

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