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General Chemistry · Stoichiometry

Stoichiometry practice problems, with worked answers

Stoichiometry is bookkeeping for reactions. A balanced equation tells you how many moles react, never how many grams. So almost every problem runs the same route: grams of what you have, to moles, through the mole ratio, to moles of what you want, then back to grams.

Already know the basics? Jump to the 8 practice questions or the worked examples.

The mole map: four steps for any mass-to-mass problem

Grams never convert straight into grams of a different substance. You always cross through moles, because the coefficients count particles, not mass.

  1. Balance the equation. Check it even when it looks balanced. Every coefficient you use later comes from this line, so a wrong coefficient ruins every step after it.
  2. Grams of given → moles of given. Divide by the molar mass of the substance you were given.
  3. Moles of given → moles of wanted. Multiply by the mole ratio from the balanced equation: (coefficient of wanted) ÷ (coefficient of given). Wanted goes on top.
  4. Moles of wanted → grams of wanted. Multiply by the molar mass of the substance you want.

grams wanted = grams given × (1 mol given ÷ Mgiven) × (b mol wanted ÷ a mol given) × (Mwanted ÷ 1 mol wanted)

Write the units on every factor. If grams of given and moles of given don't cancel cleanly, a factor is upside down. If the problem only asks for moles, stop after step 3.

Molar masses used on this page

These are standard atomic weights rounded for college General Chemistry work: H 1.008, C 12.011, N 14.007, O 15.999, Na 22.99, Cl 35.45, K 39.098, Ca 40.078, Fe 55.85 g/mol. Your textbook's table may differ in the last digit. That won't change a three-significant-figure answer.

Molar masses for the compounds on this page (g/mol)
SubstanceHow it adds upMolar mass used
H22(1.008)2.016
O22(15.999)32.00
H2O2(1.008) + 15.99918.02
CO212.011 + 2(15.999)44.01
C3H83(12.011) + 8(1.008)44.10
N22(14.007)28.01
NH314.007 + 3(1.008)17.03
NaCl22.99 + 35.4558.44
CaCO340.078 + 12.011 + 3(15.999)100.09
KClO339.098 + 35.45 + 3(15.999)122.55
Fe2O32(55.845) + 3(15.999)159.69
C6H12O66(12.011) + 12(1.008) + 6(15.999)180.16

Significant figures: keep at least one extra digit while you work, and round only the final answer. Round it to the fewest significant figures in your measured data. Coefficients and the 100 in percent yield are exact, so they don't limit anything.

Percent yield and limiting reagents

The mass you calculate with the mole map is the theoretical yield: the most product the reaction could give. What you actually collect in lab is the actual yield, and it's usually less.

percent yield = (actual yield ÷ theoretical yield) × 100%

Both yields must be in the same units, usually grams of the same product. A percent yield over 100% is a red flag: wet product, an impurity, or a math slip.

Everything here assumes one reactant is given and the others are in excess. When the problem gives amounts of two or more reactants, first find which one runs out. The limiting reagent guide walks through that step.

Worked examples

1. Balance first, then mass to mass: burning propane

Propane burns in oxygen: C3H8 + O2 → CO2 + H2O (unbalanced). How many grams of CO2 form from 22.0 g of C3H8?

  1. Balance. Three C atoms give 3 CO2. Eight H atoms give 4 H2O. Now count O on the right: 3(2) + 4(1) = 10 O atoms, so you need 5 O2. Balanced: C3H8 + 5 O2 → 3 CO2 + 4 H2O.
  2. Grams to moles: 22.0 g ÷ 44.10 g/mol = 0.4989 mol C3H8.
  3. Mole ratio: 0.4989 mol C3H8 × (3 mol CO2 ÷ 1 mol C3H8) = 1.497 mol CO2.
  4. Moles to grams: 1.497 mol × 44.01 g/mol = 65.87 g.
  5. 22.0 g has three significant figures, so round to three.

Answer: 65.9 g CO2

2. A product that is an element: iron from iron(III) oxide

In a blast furnace, Fe2O3 + 3 CO → 2 Fe + 3 CO2. What mass of iron can 50.0 g of Fe2O3 produce, with CO in excess?

  1. Check the balance: Fe 2 = 2, C 3 = 3, O 3 + 3 = 6 on the left and 3(2) = 6 on the right. Balanced.
  2. Grams to moles: 50.0 g ÷ 159.69 g/mol = 0.3131 mol Fe2O3.
  3. Mole ratio: 0.3131 mol × (2 mol Fe ÷ 1 mol Fe2O3) = 0.6262 mol Fe.
  4. Moles to grams: use the molar mass of Fe atoms (55.85 g/mol), because the product is Fe, not Fe2. 0.6262 mol × 55.85 g/mol = 34.97 g.

Answer: 35.0 g Fe

3. Percent yield: making ammonia

N2 + 3 H2 → 2 NH3. A reaction of 28.0 g of N2 with excess H2 gives 25.0 g of NH3. What is the percent yield?

  1. Theoretical yield first. Grams to moles: 28.0 g ÷ 28.01 g/mol = 0.9996 mol N2.
  2. Mole ratio: 0.9996 mol × (2 mol NH3 ÷ 1 mol N2) = 1.999 mol NH3.
  3. Moles to grams: 1.999 mol × 17.03 g/mol = 34.05 g NH3. Keep this unrounded value for the next step.
  4. Percent yield: (25.0 g ÷ 34.05 g) × 100% = 73.4%.

Answer: 73.4% yield (theoretical yield 34.0 g NH3)

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Grams-to-grams calculator, with every step in words

Enter the numbers from a balanced equation. The calculator shows the three conversions so you can compare them with your own work. It doesn't balance equations for you.

Results show four significant figures. Round your final answer to match your data.

Practice questions

Pick an answer, then check it. A wrong answer tells you what went wrong, so you can try again. Every question also has a full solution.

Question 1For C3H8 + 5 O2 → 3 CO2 + 4 H2O, how many moles of O2 are needed to burn 2.0 mol of C3H8?

Show solution

Moles to moles needs only the mole ratio. 2.0 mol C3H8 × (5 mol O2 ÷ 1 mol C3H8) = 10. mol O2. No molar masses are involved, because nothing is in grams.

Question 2Balance Al + O2 → Al2O3. What is the mole ratio of Al to Al2O3?

Show solution

Each Al2O3 has 3 O atoms, and O2 brings 2 at a time. The least common multiple of 2 and 3 is 6, so write 2 Al2O3 (6 O) and 3 O2 (6 O). Then 2 Al2O3 contain 4 Al, so you need 4 Al: 4 Al + 3 O2 → 2 Al2O3. The Al : Al2O3 ratio is 4 : 2 = 2 : 1.

Question 32 H2 + O2 → 2 H2O. How many grams of water form from 6.00 g of H2 with excess O2?

Show solution
  1. 6.00 g H2 ÷ 2.016 g/mol = 2.976 mol H2.
  2. 2.976 mol H2 × (2 mol H2O ÷ 2 mol H2) = 2.976 mol H2O.
  3. 2.976 mol × 18.02 g/mol = 53.63 g, or 53.6 g to three significant figures.
Question 42 Na + Cl2 → 2 NaCl. What mass of NaCl forms from 10.0 g of sodium with excess chlorine?

Show solution
  1. 10.0 g Na ÷ 22.99 g/mol = 0.4350 mol Na.
  2. 0.4350 mol Na × (2 mol NaCl ÷ 2 mol Na) = 0.4350 mol NaCl.
  3. 0.4350 mol × 58.44 g/mol = 25.42 g, or 25.4 g.
Question 5Heating limestone: CaCO3 → CaO + CO2. What mass of CO2 comes from 25.0 g of CaCO3?

Show solution
  1. 25.0 g ÷ 100.09 g/mol = 0.2498 mol CaCO3.
  2. The ratio is 1 : 1, so 0.2498 mol CO2.
  3. 0.2498 mol × 44.01 g/mol = 10.99 g, or 11.0 g.

Check: the CaO is 25.0 − 11.0 = 14.0 g, and 0.2498 mol × 56.08 g/mol = 14.0 g. The masses add up.

Question 6The theoretical yield of CO2 in a reaction is 11.0 g. You collect 9.35 g. What is the percent yield?

Show solution

Percent yield = (actual ÷ theoretical) × 100% = (9.35 ÷ 11.0) × 100% = 85.0%. Both values are in grams of the same product, so the units cancel.

Question 72 KClO3 → 2 KCl + 3 O2. You heat 12.25 g of KClO3, and the percent yield is 80.0%. What mass of O2 do you actually collect?

Show solution
  1. 12.25 g ÷ 122.55 g/mol = 0.09996 mol KClO3.
  2. 0.09996 mol × (3 mol O2 ÷ 2 mol KClO3) = 0.1499 mol O2.
  3. 0.1499 mol × 32.00 g/mol = 4.798 g theoretical yield.
  4. Actual = 0.800 × 4.798 g = 3.84 g.
Question 8C6H12O6 + 6 O2 → 6 CO2 + 6 H2O. How many grams of O2 are needed to burn 45.0 g of glucose completely?

Show solution
  1. 45.0 g ÷ 180.16 g/mol = 0.2498 mol C6H12O6.
  2. 0.2498 mol × (6 mol O2 ÷ 1 mol glucose) = 1.499 mol O2.
  3. 1.499 mol × 32.00 g/mol = 47.96 g, or 48.0 g.

Answer key

  1. C, 10. mol O2: 2.0 × 5/1.
  2. A, 2 : 1: 4 Al + 3 O2 → 2 Al2O3.
  3. D, 53.6 g H2O: 2.976 mol H2 gives 2.976 mol H2O.
  4. B, 25.4 g NaCl: 1:1 mole ratio of Na to NaCl.
  5. C, 11.0 g CO2: 0.2498 mol × 44.01 g/mol.
  6. A, 85.0%: 9.35 ÷ 11.0 × 100%.
  7. B, 3.84 g O2: 80.0% of the 4.80 g theoretical yield.
  8. D, 48.0 g O2: 6 mol O2 per mol glucose.

Frequently asked questions

Why can't I use the coefficients with grams?

Coefficients count particles (moles), and different substances have different masses per mole. One mole of H2 is about 2 g, while one mole of O2 is about 32 g. Convert to moles before you use the ratio.

How many significant figures should a stoichiometry answer have?

Match the fewest significant figures in the measured data, usually the given mass. Coefficients are exact counts and don't limit the answer. Carry an extra digit through the steps and round once at the end.

What if the problem gives amounts of two reactants?

Then one reactant probably runs out first. Find the limiting reagent before you calculate any product. The product amount comes from the limiting reagent, not the one in excess.

Is stoichiometry on the MCAT?

Yes. The AAMC content outline for the chemical and physical foundations section includes stoichiometry topics such as balancing equations, the mole concept, and limiting reactants. Expect calculations without a calculator, so practice estimating.

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