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(N2​)n(N2​)n(NH3​)m(NH3​)÷ M(N2​)× 2/1× M(NH3​)

For the reaction N2 + 3H2 → 2NH3, what mass m of NH3 forms from 8 g of N2? Assume the other reactants are in excess.

M(N2) = 28.02 g/molM(NH3) = 17.03 g/mol
m = g
2.Easy
m(H2​)n(H2​)n(H2​O)m(H2​O)÷ M(H2​)× 2/2× M(H2​O)

For the reaction 2H2 + O2 → 2H2O, what mass m of H2O forms from 16 g of H2? Assume the other reactants are in excess.

M(H2) = 2.016 g/molM(H2O) = 18.02 g/mol
m = g
3.Medium
m(X)n(X)n(AlCl3​)m(AlCl3​)÷ M(X)× 2/ν× M(AlCl3​)

24 g of Al and 16 g of Cl2 react according to 2Al + 3Cl2 → 2AlCl3. What mass m of AlCl3 can form?

M(Al) = 26.98M(Cl2) = 70.9M(AlCl3) = 133.33 g/mol
m = g
4.Medium
m(X)n(X)n(CO2​)m(CO2​)÷ M(X)× 3/ν× M(CO2​)

30 g of C3H8 and 24 g of O2 react according to C3H8 + 5O2 → 3CO2 + 4H2O. What mass m of CO2 can form?

M(C3H8) = 44.1M(O2) = 32M(CO2) = 44.01 g/mol
m = g
5.Hard
m(Mg)n(Mg)n(MgO)m(MgO)÷ M(Mg)× 2/2× M(MgO)

In the reaction 2Mg + O2 → 2MgO, 24 g of Mg reacts completely (the other reactants are in excess) and 29.84 g of MgO is collected. Find the percent yield Y.

M(Mg) = 24.31 g/molM(MgO) = 40.3 g/molactual: 29.84 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, 12 g of Fe2O3 reacts completely (the other reactants are in excess) and 7.39 g of Fe is collected. Find the percent yield Y.

M(Fe2O3) = 159.69 g/molM(Fe) = 55.85 g/molactual: 7.39 g
Y = %

Answer key

  1. 1. m = 9.724 g
  2. 2. m = 143 g
  3. 3. m = 20.06 g
  4. 4. m = 19.8 g
  5. 5. Y = 75 %
  6. 6. Y = 88.04 %

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