Stoichiometry — Chemistry Practice Worksheet

Grades 10–11 Hard difficulty 10 problems

Stoichiometry uses balanced chemical equations to relate amounts of reactants and products. Coefficients in the equation represent mole ratios, not grams or molecules directly.

The mole bridge connects mass (grams) to moles via molar mass (g/mol), and moles to particles via Avogadro's number (6.02×10236.02 \times 10^{23}). Always convert to moles first for ratio calculations.

The limiting reagent is the reactant that runs out first and determines maximum product. Excess reagent remains after the reaction stops.

Skills practiced

  • Balancing chemical equations
  • Mole-to-mole conversions
  • Mass-to-mole stoichiometry
  • Identifying limiting reagents
Name: Date: Period:

Practice Worksheet: Stoichiometry

Instructions: Solve each problem carefully. Show all work clearly. Write your final answer in the space provided or on a separate sheet as directed.

  1. 1.How many grams of oxygen gas are required to completely react with 25.0 grams of hydrogen gas to produce water? The balanced equation is: 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}
  2. 2.Consider the reaction: N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 If 15.0 grams of nitrogen gas react with excess hydrogen, what mass of ammonia (in grams) is produced?
  3. 3.Iron(III) oxide reacts with carbon monoxide to produce iron and carbon dioxide: Fe2O3+3CO→2Fe+3CO2\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2 How many moles of carbon monoxide are needed to produce 112 grams of iron?
  4. 4.A student reacts 50.0 grams of calcium carbonate (CaCO3\text{CaCO}_3) with excess hydrochloric acid: CaCO3+2HCl→CaCl2+H2O+CO2\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 What volume of carbon dioxide gas (at STP) is produced? (At STP, 1 mole of gas occupies 22.4 L.)
  5. 5.In the reaction: 2Al+3Cl2→2AlCl32\text{Al} + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_3 If 10.0 grams of aluminum react with 35.0 grams of chlorine gas, which reactant is limiting? What mass of aluminum chloride is produced?
  6. 6.How many molecules of water are produced when 5.00 grams of methane (CH4\text{CH}_4) are burned completely in oxygen? CH4+2O2→CO2+2H2O\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}
  7. 7.For the reaction: 4NH3+5O2→4NO+6H2O4\text{NH}_3 + 5\text{O}_2 \rightarrow 4\text{NO} + 6\text{H}_2\text{O} Calculate the mass of nitrogen monoxide produced when 34.0 grams of ammonia react with 80.0 grams of oxygen. Identify the limiting reactant.
  8. 8.A sample of 25.0 grams of sodium hydroxide (NaOH\text{NaOH}) is neutralized by sulfuric acid (H2SO4\text{H}_2\text{SO}_4): 2NaOH+H2SO4→Na2SO4+2H2O2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} What mass of sodium sulfate is produced?
  9. 9.In the reaction: Zn+2HCl→ZnCl2+H2\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 If 13.0 grams of zinc react with excess hydrochloric acid, what volume of hydrogen gas (at STP) is produced?
  10. 10.Consider the combustion of propane: C3H8+5O2→3CO2+4H2O\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} If 44.0 grams of propane are burned, how many grams of carbon dioxide are produced? How many grams of oxygen are consumed?

Answer Key

  1. 1.
    Molar mass H2_2 = 2.02 g/mol; moles H2_2 = 25.0 / 2.02 = 12.38 mol; mole ratio O2_2:H2_2 = 1:2, so moles O2_2 = 6.19 mol; mass O2_2 = 6.19 ×\times 32.00 = 198.1 g ≈\approx 199 g.
    Final answer: 199 g O2_2
  2. 2.
    Molar mass N2_2 = 28.02 g/mol; moles N2_2 = 15.0 / 28.02 = 0.5353 mol; mole ratio NH3_3:N2_2 = 2:1, so moles NH3_3 = 1.0706 mol; mass NH3_3 = 1.0706 ×\times 17.04 = 18.24 g ≈\approx 18.2 g.
    Final answer: 18.2 g NH3_3
  3. 3.
    Molar mass Fe = 55.85 g/mol; moles Fe = 112 / 55.85 = 2.005 mol; mole ratio CO:Fe = 3:2, so moles CO = (3/2) ×\times 2.005 = 3.008 mol ≈\approx 3.01 mol.
    Final answer: 3.01 mol CO
  4. 4.
    Molar mass CaCO3_3 = 100.09 g/mol; moles CaCO3_3 = 50.0 / 100.09 = 0.4996 mol; mole ratio CO2_2:CaCO3_3 = 1:1, so moles CO2_2 = 0.4996 mol; volume at STP = 0.4996 ×\times 22.4 = 11.19 L ≈\approx 11.2 L.
    Final answer: 11.2 L CO2_2
  5. 5.
    Molar mass Al = 26.98 g/mol; moles Al = 10.0 / 26.98 = 0.3706 mol; molar mass Cl2_2 = 70.90 g/mol; moles Cl2_2 = 35.0 / 70.90 = 0.4937 mol; mole ratio Al:Cl2_2 = 2:3; required Cl2_2 for Al = (3/2) ×\times 0.3706 = 0.5559 mol (more than available, so Cl2_2 limiting); moles AlCl3_3 from Cl2_2 = (2/3) ×\times 0.4937 = 0.3291 mol; mass AlCl3_3 = 0.3291 ×\times 133.34 = 43.88 g ≈\approx 43.8 g.
    Final answer: Cl2_2 is limiting; 43.8 g AlCl3_3
  6. 6.
    Molar mass CH4_4 = 16.04 g/mol; moles CH4_4 = 5.00 / 16.04 = 0.3117 mol; mole ratio H2_2O:CH4_4 = 2:1, so moles H2_2O = 0.6234 mol; molecules = 0.6234 ×\times 6.022 ×\times 1023^{23} = 3.754×10233.754 \times 10^{23} ≈\approx 3.76×10233.76 \times 10^{23}.
    Final answer: 3.76×10233.76 \times 10^{23} molecules H2_2O
  7. 7.
    Molar mass NH3_3 = 17.03 g/mol; moles NH3_3 = 34.0 / 17.03 = 1.997 mol; molar mass O2_2 = 32.00 g/mol; moles O2_2 = 80.0 / 32.00 = 2.50 mol; mole ratio NH3_3:O2_2 = 4:5; required O2_2 for NH3_3 = (5/4) ×\times 1.997 = 2.496 mol (available 2.50 mol, so NH3_3 slightly in excess? Check: required NH3_3 for O2_2 = (4/5) ×\times 2.50 = 2.00 mol, available 1.997 mol, so NH3_3 is limiting? Let's recalc: 1.997 mol NH3_3 needs (5/4)×\times1.997 = 2.496 mol O2_2, we have 2.50 mol O2_2, so O2_2 is in excess, NH3_3 is limiting. Moles NO = moles NH3_3 = 1.997 mol; mass NO = 1.997 ×\times 30.01 = 59.93 g ≈\approx 60.0 g. (Alternatively, using O2_2: 2.50 mol O2_2 would produce (4/5)×\times2.50 = 2.00 mol NO, but NH3_3 only gives 1.997 mol, so NH3_3 is limiting.)
    Final answer: O2_2 is limiting; 60.0 g NO
  8. 8.
    Molar mass NaOH = 40.00 g/mol; moles NaOH = 25.0 / 40.00 = 0.625 mol; mole ratio Na2_2SO4_4:NaOH = 1:2, so moles Na2_2SO4_4 = 0.3125 mol; molar mass Na2_2SO4_4 = 142.04 g/mol; mass = 0.3125 ×\times 142.04 = 44.39 g ≈\approx 44.4 g.
    Final answer: 44.4 g Na2_2SO4_4
  9. 9.
    Molar mass Zn = 65.38 g/mol; moles Zn = 13.0 / 65.38 = 0.1988 mol; mole ratio H2_2:Zn = 1:1, so moles H2_2 = 0.1988 mol; volume at STP = 0.1988 ×\times 22.4 = 4.453 L ≈\approx 4.46 L.
    Final answer: 4.46 L H2_2
  10. 10.
    Molar mass C3_3H8_8 = 44.10 g/mol; moles C3_3H8_8 = 44.0 / 44.10 = 0.9977 mol; mole ratio CO2_2:C3_3H8_8 = 3:1, so moles CO2_2 = 2.993 mol; mass CO2_2 = 2.993 ×\times 44.01 = 131.7 g ≈\approx 132 g. Mole ratio O2_2:C3_3H8_8 = 5:1, so moles O2_2 = 4.989 mol; mass O2_2 = 4.989 ×\times 32.00 = 159.6 g ≈\approx 160 g.
    Final answer: 132 g CO2_2; 160 g O2_2
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Common mistakes to avoid

  • Using unbalanced equations for mole ratios
  • Confusing molar mass with molecular mass units
  • Ignoring limiting reagent in multi-reactant problems
  • Rounding moles too early in multi-step problems

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