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.

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(Fe2​O3​)n(Fe2​O3​)n(CO2​)m(CO2​)÷ M(Fe2​O3​)× 3/1× M(CO2​)

For the reaction Fe2O3 + 3CO → 2Fe + 3CO2, what mass m of CO2 forms from 5 g of Fe2O3? Assume the other reactants are in excess.

M(Fe2O3) = 159.69 g/molM(CO2) = 44.01 g/mol
m = g
2.Easy
m(Na)n(Na)n(NaCl)m(NaCl)÷ M(Na)× 2/2× M(NaCl)

For the reaction 2Na + Cl2 → 2NaCl, what mass m of NaCl forms from 100 g of Na? Assume the other reactants are in excess.

M(Na) = 22.99 g/molM(NaCl) = 58.44 g/mol
m = g
3.Medium
m(X)n(X)n(H2​O)m(H2​O)÷ M(X)× 2/ν× M(H2​O)

24 g of H2 and 16 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(NaCl)m(NaCl)÷ M(X)× 2/ν× M(NaCl)

24 g of Na and 10 g of Cl2 react according to 2Na + Cl2 → 2NaCl. What mass m of NaCl can form?

M(Na) = 22.99M(Cl2) = 70.9M(NaCl) = 58.44 g/mol
m = g
5.Hard
m(H2​)n(H2​)n(H2​O)m(H2​O)÷ M(H2​)× 2/2× M(H2​O)

In the reaction 2H2 + O2 → 2H2O, 4 g of H2 reacts completely (the other reactants are in excess) and 27.89 g of H2O is collected. Find the percent yield Y.

M(H2) = 2.016 g/molM(H2O) = 18.02 g/molactual: 27.89 g
Y = %
6.Hard
m(C6​H12​O6​)n(C6​H12​O6​)n(CO2​)m(CO2​)÷ M(C6​H12​O6​)× 2/1× M(CO2​)

In the reaction C6H12O6 → 2C2H5OH + 2CO2, 24 g of C6H12O6 reacts completely (the other reactants are in excess) and 11.02 g of CO2 is collected. Find the percent yield Y.

M(C6H12O6) = 180.16 g/molM(CO2) = 44.01 g/molactual: 11.02 g
Y = %

Answer key

  1. 1. m = 4.134 g
  2. 2. m = 254.2 g
  3. 3. m = 18.02 g
  4. 4. m = 16.49 g
  5. 5. Y = 78.01 %
  6. 6. Y = 93.98 %

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