Stoichiometry - AP Chemistry

1. Sulfuric acid neutralizes sodium hydroxide according to the balanced equation below: $\text{H}_2\text{SO}_4(aq)+2\,\text{NaOH}(aq)\rightarrow \text{Na}_2\text{SO}_4(aq)+2\,\text{H}_2\text{O}(l)$ If $30.0\ \text{mL}$ of $0.50\ \text{M}$ $\text{H}_2\text{SO}_4(aq)$ reacts completely with excess $\text{NaOH}(aq)$, how many moles of $\text{Na}_2\text{SO}_4(aq)$ are formed?
   Answer: $0.015\ \text{mol}$ This stoichiometry problem combines molarity with acid-base neutralization. First, calculate moles of H₂SO₄: (0.0300 L) × (0.50 mol/L) = 0.015 mol H₂SO₄. The balanced equation shows a 1:1 ratio between H₂SO₄ and Na₂SO₄, so 0.015 mol H₂SO₄ produces 0.015 mol Na₂SO₄. Choice A (0.030 mol) might come from using the coefficient 2 from NaOH incorrectly. Always focus on the mole ratio between your given reactant and desired product, ignoring other species' coefficients.
2. Hydrogen gas reacts with oxygen gas to form water according to the balanced equation $2\text{H}_2(g)+\text{O}_2(g)\rightarrow 2\text{H}_2\text{O}(l)$. If $3.0\ \text{mol}$ of $\text{O}_2$ reacts completely with excess $\text{H}_2$, how many moles of $\text{H}_2\text{O}$ are produced?
   Answer: $6.0\ \text{mol}$ This question tests stoichiometry. The balanced equation 2H₂(g) + O₂(g) → 2H₂O(l) provides mole ratios, such as 1 mol O₂ to 2 mol H₂O. These ratios act as conversion factors to relate the given amount of reactant to the desired product. For 3.0 mol O₂, the calculation is 3.0 mol O₂ × (2 mol H₂O / 1 mol O₂) = 6.0 mol H₂O. A tempting distractor is 3.0 mol, which occurs if one incorrectly assumes a 1:1 ratio instead of 2:1. Always start from the balanced equation and convert to moles before applying ratios.
3. Iron reacts with oxygen to form iron(III) oxide: $\mathrm{4\,Fe(s) + 3\,O_2(g) \rightarrow 2\,Fe_2O_3(s)}$ If $0.75\ \text{mol}$ of $\mathrm{O_2}$ reacts completely with excess $\mathrm{Fe}$, how many moles of $\mathrm{Fe_2O_3}$ are produced?
   Answer: $0.50\ \text{mol}$ 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.
4. The reaction of hydrogen gas with oxygen gas forms water according to the balanced equation below: $2\,\text{H}_2(g)+\text{O}_2(g)\rightarrow 2\,\text{H}_2\text{O}(l)$ If $3.0\ \text{mol}$ of $\text{O}_2(g)$ reacts with excess $\text{H}_2(g)$, how many moles of $\text{H}_2\text{O}(l)$ are produced?
   Answer: $6.0\ \text{mol}$ 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.
5. Sulfuric acid reacts with potassium hydroxide according to the balanced equation $\mathrm{H_2SO_4(aq) + 2KOH(aq) \rightarrow K_2SO_4(aq) + 2H_2O(l)}$. If $0.150\ \text{mol}$ of $\mathrm{KOH}$ reacts completely with excess $\mathrm{H_2SO_4}$, how many moles of $\mathrm{H_2O}$ are produced?
   Answer: $0.150\ \text{mol}$ This problem requires stoichiometry to find moles of water produced in an acid-base reaction. The balanced equation H₂SO₄(aq) + 2KOH(aq) → K₂SO₄(aq) + 2H₂O(l) shows that 2 moles of KOH produce 2 moles of H₂O, giving a 1:1 ratio. Starting with 0.150 mol KOH, we calculate: 0.150 mol KOH × (2 mol H₂O)/(2 mol KOH) = 0.150 mol H₂O. A mistake would be to think that since H₂SO₄ has a 1:2 ratio with H₂O, we should divide 0.150 by 2 to get 0.0750 mol (answer B), but this ignores that we're given KOH, not H₂SO₄. Always identify which reactant you're given and use its specific mole ratio with the desired product.
6. Aluminum reacts with chlorine gas to form aluminum chloride according to the balanced equation below. $2\,\text{Al}(s)+3\,\text{Cl}_2(g)\rightarrow 2\,\text{AlCl}_3(s)$ If $0.60\ \text{mol}$ of $\text{Cl}_2(g)$ reacts with excess $\text{Al}(s)$, how many moles of $\text{AlCl}_3(s)$ can form?
   Answer: $0.40\ \text{mol}$ This problem requires stoichiometry to determine moles of AlCl₃ produced from Cl₂. The balanced equation 2Al(s) + 3Cl₂(g) → 2AlCl₃(s) shows that 3 moles of Cl₂ produce 2 moles of AlCl₃, giving a 3:2 ratio. With 0.60 mol Cl₂ and excess Al, we calculate: 0.60 mol Cl₂ × (2 mol AlCl₃/3 mol Cl₂) = 0.40 mol AlCl₃. A student might incorrectly use a 1:1 ratio and conclude that 0.60 mol Cl₂ produces 0.60 mol AlCl₃, which would be answer C. The correct approach is to use the exact coefficients from the balanced equation to establish the proper mole ratio between reactants and products.
7. Hydrogen gas reacts with chlorine gas according to the balanced equation: $\;\text{H}_2(g)+\text{Cl}_2(g)\rightarrow 2\text{HCl}(g)$. If $0.150\ \text{mol}$ of $\text{H}_2$ reacts completely with excess $\text{Cl}_2$, how many moles of $\text{HCl}$ are formed?
   Answer: $0.300\ \text{mol}$ 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.
8. Calcium carbonate reacts with hydrochloric acid according to the balanced equation below: $\mathrm{CaCO_3(s) + 2\,HCl(aq) \rightarrow CaCl_2(aq) + CO_2(g) + H_2O(l)}$ If $0.30\ \text{mol}$ of $\mathrm{CaCO_3}$ reacts completely with excess $\mathrm{HCl}$, how many moles of $\mathrm{HCl}$ are consumed?
   Answer: $0.60\ \text{mol}$ 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.
9. Propane combusts according to the balanced equation $\mathrm{C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)}$. If $0.20\ \text{mol}$ of $\mathrm{C_3H_8}$ combusts completely with excess $\mathrm{O_2}$, how many moles of $\mathrm{CO_2}$ are produced?
   Answer: $0.60\ \text{mol}$ This problem involves stoichiometry in a combustion reaction to find moles of CO₂ produced. The balanced equation C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l) shows that 1 mole of propane produces 3 moles of CO₂. Starting with 0.20 mol C₃H₈, we calculate: 0.20 mol C₃H₈ × (3 mol CO₂)/(1 mol C₃H₈) = 0.60 mol CO₂. A tempting mistake would be to use the oxygen coefficient of 5, perhaps calculating 0.20 × 5 = 1.00 mol (answer E), which confuses the relationships between different species. Always identify the specific mole ratio between your given reactant and desired product from the balanced equation, ignoring coefficients of other species.
10. Sodium bicarbonate reacts with hydrochloric acid according to the balanced equation below: $\text{NaHCO}_3(aq)+\text{HCl}(aq)\rightarrow \text{NaCl}(aq)+\text{H}_2\text{O}(l)+\text{CO}_2(g)$ If $25.0\ \text{mL}$ of $0.200\ \text{M}$ $\text{NaHCO}_3(aq)$ reacts with excess $\text{HCl}(aq)$, how many moles of $\text{CO}_2(g)$ are produced?
    Answer: $0.00500\ \text{mol}$ This stoichiometry problem combines molarity calculations with mole ratios. First, calculate moles of NaHCO₃: (0.0250 L) × (0.200 mol/L) = 0.00500 mol NaHCO₃. The balanced equation shows a 1:1 ratio between NaHCO₃ and CO₂, so 0.00500 mol NaHCO₃ produces 0.00500 mol CO₂. Choice C (0.200 mol) incorrectly uses the molarity value directly without considering volume. Always convert volume and molarity to moles before applying stoichiometric ratios.
11. Calcium hydroxide reacts with phosphoric acid according to the balanced equation below: $3\,\text{Ca(OH)}_2(aq)+2\,\text{H}_3\text{PO}_4(aq)\rightarrow \text{Ca}_3(\text{PO}_4)_2(s)+6\,\text{H}_2\text{O}(l)$ If $0.30\ \text{mol}$ of $\text{Ca(OH)}_2(aq)$ reacts with excess $\text{H}_3\text{PO}_4(aq)$, how many moles of $\text{Ca}_3(\text{PO}_4)_2(s)$ can form?
    Answer: $0.10\ \text{mol}$ This stoichiometry problem involves a complex acid-base reaction. The balanced equation shows that 3 mol Ca(OH)₂ produces 1 mol Ca₃(PO₄)₂, giving a 3:1 ratio. With 0.30 mol Ca(OH)₂: (0.30 mol Ca(OH)₂) × (1 mol Ca₃(PO₄)₂/3 mol Ca(OH)₂) = 0.10 mol Ca₃(PO₄)₂. Choice C (0.30 mol) incorrectly assumes a 1:1 ratio between reactant and product. Always use the coefficients from the balanced equation, even when they create non-intuitive ratios.
12. Potassium chlorate decomposes upon heating according to the balanced equation below: $2\,\text{KClO}_3(s)\rightarrow 2\,\text{KCl}(s)+3\,\text{O}_2(g)$ If $0.40\ \text{mol}$ of $\text{KClO}_3(s)$ decomposes completely, how many moles of $\text{O}_2(g)$ are produced?
    Answer: $0.60\ \text{mol}$ This stoichiometry problem involves the decomposition of potassium chlorate. The balanced equation shows that 2 mol KClO₃ produces 3 mol O₂, giving a 2:3 mole ratio. With 0.40 mol KClO₃, we calculate: (0.40 mol KClO₃) × (3 mol O₂/2 mol KClO₃) = 0.60 mol O₂. Choice D (0.40 mol) incorrectly assumes a 1:1 ratio, ignoring the coefficients. Always use the exact coefficients from the balanced equation as your mole ratio for conversions.
13. Hydrogen peroxide decomposes according to the balanced equation $\mathrm{2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)}$. If $0.80\ \text{mol}$ of $\mathrm{H_2O_2}$ decomposes completely, how many moles of $\mathrm{O_2}$ are produced?
    Answer: $0.40\ \text{mol}$ 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.
14. Sulfuric acid reacts with sodium hydroxide according to the balanced equation below. $\text{H}_2\text{SO}_4(aq)+2\,\text{NaOH}(aq)\rightarrow \text{Na}_2\text{SO}_4(aq)+2\,\text{H}_2\text{O}(\ell)$ If $0.30\ \text{mol}$ of $\text{NaOH}(aq)$ reacts with excess $\text{H}_2\text{SO}_4(aq)$, how many moles of $\text{Na}_2\text{SO}_4(aq)$ can be produced?
    Answer: $0.15\ \text{mol}$ This problem involves stoichiometry to find moles of Na₂SO₄ produced from NaOH. The balanced equation H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(ℓ) shows that 2 moles of NaOH produce 1 mole of Na₂SO₄, a 2:1 ratio. Starting with 0.30 mol NaOH and excess H₂SO₄, we calculate: 0.30 mol NaOH × (1 mol Na₂SO₄/2 mol NaOH) = 0.15 mol Na₂SO₄. A common error would be to assume a 1:1 ratio and conclude that 0.30 mol NaOH produces 0.30 mol Na₂SO₄, giving answer C. The key strategy is to carefully examine the balanced equation coefficients to establish the correct mole ratio between reactants and products.
15. Magnesium reacts with oxygen according to the balanced equation $\mathrm{2Mg(s) + O_2(g) \rightarrow 2MgO(s)}$. If $1.00\ \text{mol}$ of $\mathrm{Mg}$ reacts with excess $\mathrm{O_2}$, how many moles of $\mathrm{O_2}$ are consumed?
    Answer: $0.50\ \text{mol}$ 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.
16. Calcium carbonate decomposes when heated according to the balanced equation below. $\text{CaCO}_3(s)\rightarrow \text{CaO}(s)+\text{CO}_2(g)$ If $10.0\ \text{g}$ of $\text{CaCO}_3(s)$ (molar mass $=100.1\ \text{g mol}^{-1}$) decomposes completely, what mass of $\text{CO}_2(g)$ is produced? (Molar mass of $\text{CO}_2=44.0\ \text{g mol}^{-1}$.)
    Answer: $4.4\ \text{g}$ This problem uses stoichiometry to calculate the mass of CO₂ produced from CaCO₃ decomposition. The balanced equation CaCO₃(s) → CaO(s) + CO₂(g) shows a 1:1 mole ratio between CaCO₃ and CO₂. First, convert 10.0 g CaCO₃ to moles: 10.0 g ÷ 100.1 g/mol = 0.0999 mol CaCO₃. Since the ratio is 1:1, this produces 0.0999 mol CO₂. Converting to mass: 0.0999 mol × 44.0 g/mol = 4.4 g CO₂. A common error would be to assume that 10.0 g of reactant produces 10.0 g of product, ignoring the need for mole conversions. The key strategy is to always convert to moles first, apply the stoichiometric ratio, then convert back to the desired units.
17. Aqueous silver nitrate reacts with aqueous sodium chloride according to the balanced equation $\mathrm{AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_3(aq)}$. If $50.0\ \text{mL}$ of $0.100\ \text{M}$ $\mathrm{AgNO_3}$ is mixed with excess $\mathrm{NaCl}$, how many moles of $\mathrm{AgCl}$ precipitate form?
    Answer: $0.00500\ \text{mol}$ This problem requires stoichiometry to find moles of silver chloride precipitate formed. The balanced equation AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq) shows a 1:1 mole ratio between AgNO₃ and AgCl. First, calculate moles of AgNO₃: 0.0500 L × 0.100 mol/L = 0.00500 mol AgNO₃. Since the ratio is 1:1, this produces 0.00500 mol AgCl. A common error would be to use 50.0 mL directly without converting to liters, perhaps calculating 50.0 × 0.100 = 5.00, then selecting answer E (0.0500 mol) thinking it's already in the correct units. Always convert volume to liters when working with molarity, then apply the stoichiometric ratios from the balanced equation.
18. Magnesium reacts with nitrogen gas to form magnesium nitride: $\mathrm{3\,Mg(s) + N_2(g) \rightarrow Mg_3N_2(s)}$ If $0.90\ \text{mol}$ of $\mathrm{Mg}$ reacts completely with excess $\mathrm{N_2}$, how many moles of $\mathrm{Mg_3N_2}$ are formed?
    Answer: $0.30\ \text{mol}$ 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.
19. Copper reacts with nitric acid to produce copper(II) nitrate according to the balanced equation below: $\text{Cu}(s)+4\,\text{HNO}_3(aq)\rightarrow \text{Cu(NO}_3)_2(aq)+2\,\text{NO}_2(g)+2\,\text{H}_2\text{O}(l)$ If $0.80\ \text{mol}$ of $\text{HNO}_3(aq)$ reacts with excess $\text{Cu}(s)$, how many moles of $\text{NO}_2(g)$ are produced?
    Answer: $0.40\ \text{mol}$ This stoichiometry problem asks for NO₂ produced from HNO₃. The balanced equation shows that $4 \text{ mol HNO}_3$ produces $2 \text{ mol NO}_2$, establishing a $4:2$ (or $2:1$) ratio. Starting with $0.80 \text{ mol HNO}_3$: ($0.80 \text{ mol HNO}_3$) × ($\frac{2 \text{ mol NO}_2}{4 \text{ mol HNO}_3}$) = $0.40 \text{ mol NO}_2$. Choice B ($0.20 \text{ mol}$) might result from using a 4:1 ratio instead of 4:2. Always simplify ratios when possible, but ensure you're using the correct coefficients from the balanced equation.
20. Ammonia is produced by the Haber process: $\mathrm{N_2(g) + 3\,H_2(g) \rightarrow 2\,NH_3(g)}$ If $0.50\ \text{mol}$ of $\mathrm{N_2}$ reacts completely with excess $\mathrm{H_2}$, how many moles of $\mathrm{NH_3}$ can be produced?
    Answer: $1.0\ \text{mol}$ This problem involves stoichiometry to calculate ammonia production in the Haber process. The balanced equation N₂ + 3H₂ → 2NH₃ reveals that 1 mole of N₂ produces 2 moles of NH₃, establishing a 2:1 mole ratio. Starting with 0.50 mol N₂, we calculate: 0.50 mol N₂ × (2 mol NH₃/1 mol N₂) = 1.0 mol NH₃. A tempting error would be to divide by 2 instead of multiply, giving 0.25 mol, which incorrectly reverses the mole ratio. Remember to always set up the mole ratio as a fraction with the desired substance in the numerator and the given substance in the denominator.

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