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(C6​H12​O6​)n(C6​H12​O6​)n(CO2​)m(CO2​)÷ M(C6​H12​O6​)× 2/1× M(CO2​)

For the reaction C6H12O6 → 2C2H5OH + 2CO2, what mass m of CO2 forms from 50 g of C6H12O6? Assume the other reactants are in excess.

M(C6H12O6) = 180.16 g/molM(CO2) = 44.01 g/mol
m = g
2.Easy
m(CaCO3​)n(CaCO3​)n(CO2​)m(CO2​)÷ M(CaCO3​)× 1/1× M(CO2​)

For the reaction CaCO3 → CaO + CO2, what mass m of CO2 forms from 50 g of CaCO3? Assume the other reactants are in excess.

M(CaCO3) = 100.09 g/molM(CO2) = 44.01 g/mol
m = g
3.Medium
m(X)n(X)n(H2​O)m(H2​O)÷ M(X)× 2/ν× M(H2​O)

20 g of H2 and 64 g of O2 react according to 2H2 + O2 → 2H2O. What mass m of H2O can form?

M(H2) = 2.016M(O2) = 32M(H2O) = 18.02 g/mol
m = g
4.Medium
m(X)n(X)n(H2​O)m(H2​O)÷ M(X)× 4/ν× M(H2​O)

12 g of C3H8 and 24 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(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 5.2 g of Fe is collected. Find the percent yield Y.

M(Fe2O3) = 159.69 g/molM(Fe) = 55.85 g/molactual: 5.2 g
Y = %
6.Hard
m(CH4​)n(CH4​)n(H2​O)m(H2​O)÷ M(CH4​)× 2/1× M(H2​O)

In the reaction CH4 + 2O2 → CO2 + 2H2O, 30 g of CH4 reacts completely (the other reactants are in excess) and 57.3 g of H2O is collected. Find the percent yield Y.

M(CH4) = 16.04 g/molM(H2O) = 18.02 g/molactual: 57.3 g
Y = %

Answer key

  1. 1. m = 24.43 g
  2. 2. m = 21.99 g
  3. 3. m = 72.08 g
  4. 4. m = 10.81 g
  5. 5. Y = 61.95 %
  6. 6. Y = 85.01 %

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