Stoichiometry worksheet with answers

Free printable Stoichiometry worksheet: 6 problems from easy to exam level, with an answer key. Print it or save it as a PDF.

New worksheetPractise online with hints

For teachers: free to print, copy and use in class. Every new worksheet has new values and its own answer key. A link to ExamProblems on your course page is appreciated.

Name: ____________________   Date: ____________

1.Easy
m(Mg)n(Mg)n(MgO)m(MgO)÷ M(Mg)× 2/2× M(MgO)

For the reaction 2Mg + O2 → 2MgO, what mass m of MgO forms from 32 g of Mg? Assume the other reactants are in excess.

M(Mg) = 24.31 g/molM(MgO) = 40.3 g/mol
m = g
2.Easy
m(C3​H8​)n(C3​H8​)n(H2​O)m(H2​O)÷ M(C3​H8​)× 4/1× M(H2​O)

For the reaction C3H8 + 5O2 → 3CO2 + 4H2O, what mass m of H2O forms from 32 g of C3H8? Assume the other reactants are in excess.

M(C3H8) = 44.1 g/molM(H2O) = 18.02 g/mol
m = g
3.Medium
m(X)n(X)n(MgO)m(MgO)÷ M(X)× 2/ν× M(MgO)

20 g of Mg and 32 g of O2 react according to 2Mg + O2 → 2MgO. What mass m of MgO can form?

M(Mg) = 24.31M(O2) = 32M(MgO) = 40.3 g/mol
m = g
4.Medium
m(X)n(X)n(H2​O)m(H2​O)÷ M(X)× 4/ν× M(H2​O)

30 g of C3H8 and 48 g of O2 react according to C3H8 + 5O2 → 3CO2 + 4H2O. What mass m of H2O can form?

M(C3H8) = 44.1M(O2) = 32M(H2O) = 18.02 g/mol
m = g
5.Hard
m(Al)n(Al)n(AlCl3​)m(AlCl3​)÷ M(Al)× 2/2× M(AlCl3​)

In the reaction 2Al + 3Cl2 → 2AlCl3, 30 g of Al reacts completely (the other reactants are in excess) and 134.9 g of AlCl3 is collected. Find the percent yield Y.

M(Al) = 26.98 g/molM(AlCl3) = 133.33 g/molactual: 134.9 g
Y = %
6.Hard
m(Fe2​O3​)n(Fe2​O3​)n(Fe)m(Fe)÷ M(Fe2​O3​)× 2/1× M(Fe)

In the reaction Fe2O3 + 3CO → 2Fe + 3CO2, 16 g of Fe2O3 reacts completely (the other reactants are in excess) and 6.94 g of Fe is collected. Find the percent yield Y.

M(Fe2O3) = 159.69 g/molM(Fe) = 55.85 g/molactual: 6.94 g
Y = %

Answer key

  1. 1. m = 53.05 g
  2. 2. m = 52.3 g
  3. 3. m = 33.16 g
  4. 4. m = 21.62 g
  5. 5. Y = 91 %
  6. 6. Y = 62.01 %

Full step-by-step solutions: practise this topic on examproblems.com.