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Chemistry 20

Chemistry 20 · Worksheets

Stoichiometry

Ten questions of mixed difficulty, covering Stoichiometry. Print it, or work through it on screen — the answer key starts on its own page.

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Stoichiometry

Chemistry 20 · maddyhelps.com

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Circle the best answer for each question. Show your work in the space provided.

  1. In 4Al + 3O₂ → 2Al₂O₃, the amount of O₂ needed to react with 2.0 mol of Al is

    1. a) 2.0 mol
    2. b) 0.67 mol
    3. c) 3.0 mol
    4. d) 1.5 mol

  2. Burning 1.00 mol of CH₄ completely produces a mass of CO₂ of about

    1. a) 88.0 g
    2. b) 22.0 g
    3. c) 16.0 g
    4. d) 44.0 g

  3. When 2.0 mol of H₂ reacts with 1.5 mol of O₂ by 2H₂ + O₂ → 2H₂O, the amount of water formed is

    1. a) 3.5 mol
    2. b) 2.0 mol, with H₂ limiting
    3. c) 3.0 mol, with O₂ limiting
    4. d) 1.5 mol, with O₂ limiting

  4. The amount of solute in 50.0 mL of 0.200 mol/L solution is

    1. a) 10.0 mol
    2. b) 0.200 mol
    3. c) 0.0100 mol
    4. d) 4.00 mol

  5. Decomposing 0.500 mol of CaCO₃ by CaCO₃ → CaO + CO₂ releases a volume of CO₂ at SATP of

    1. a) 12.4 L
    2. b) 0.500 L
    3. c) 24.8 L
    4. d) 11.2 L

  6. The mass of 0.250 mol of NaCl is about

    1. a) 58.4 g
    2. b) 0.250 g
    3. c) 14.6 g
    4. d) 234 g

  7. In 2H₂ + O₂ → 2H₂O, the mole ratio of H₂ to H₂O is

    1. a) 1:1
    2. b) 1:2
    3. c) 2:1
    4. d) 2:3

  8. The molar mass of water is about

    1. a) 18.02 g/mol
    2. b) 36.0 g/mol
    3. c) 10.0 g/mol
    4. d) 2.02 g/mol

  9. One mole of any substance contains

    1. a) 6.02 × 10²³ particles
    2. b) 12 particles
    3. c) 1 gram of particles
    4. d) 6.02 × 10²³ grams

  10. A balanced chemical equation gives the ratio in which substances react, measured in

    1. a) moles
    2. b) grams
    3. c) particles per gram
    4. d) litres

Answer key · Stoichiometry

Chemistry 20 · maddyhelps.com

  1. d) 1.5 mol — The ratio is 4 Al to 3 O₂, so O₂ = 2.0 × 3/4 = 1.5 mol. Setting the ratio up as a fraction with the wanted substance on top keeps it the right way up.
  2. d) 44.0 g — The equation is CH₄ + 2O₂ → CO₂ + 2H₂O, a 1:1 ratio, so 1.00 mol of CO₂ forms. Its mass is (1.00)(44.01) ≈ 44.0 g. Balancing first is what tells you the ratio is 1:1.
  3. b) 2.0 mol, with H₂ limiting — 2.0 mol of H₂ needs only 1.0 mol of O₂, and 1.5 mol is available, so hydrogen runs out first. The 1:1 ratio of H₂ to H₂O then gives 2.0 mol of water, and 0.5 mol of O₂ is left over.
  4. c) 0.0100 mol — n = cV = (0.200)(0.0500) = 0.0100 mol. Converting millilitres to litres first is the step most often skipped, and it costs a factor of a thousand.
  5. a) 12.4 L — The ratio is 1:1, so 0.500 mol of CO₂ forms, and V = nV_m = (0.500)(24.8) = 12.4 L. Using 22.4 L/mol here would be answering for STP, which is a different set of conditions.
  6. c) 14.6 g — m = nM = (0.250)(58.44) ≈ 14.6 g. A quarter of a mole should weigh about a quarter of the molar mass, which is a quick way to check the arithmetic went the right way.
  7. a) 1:1 — The coefficients are both 2, which reduces to 1:1. Every two hydrogen molecules give two water molecules, so the amounts match even though the substances are different.
  8. a) 18.02 g/mol — Add the atomic masses: 2(1.01) + 16.00 = 18.02 g/mol. Molar mass is read straight off the periodic table, so it is the one number in this unit you never have to derive.
  9. a) 6.02 × 10²³ particles — Avogadro's number is a count, not a mass. A mole of carbon and a mole of lead contain the same number of atoms but weigh very different amounts.
  10. a) moles — Coefficients count particles, and a mole is just a large fixed count, so the coefficients are mole ratios. They are not mass ratios — 2H₂ + O₂ → 2H₂O is 2:1 in moles but 1:8 in grams.