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(Al)n(Al)n(AlCl3​)m(AlCl3​)÷ M(Al)× 2/2× M(AlCl3​)

For the reaction 2Al + 3Cl2 → 2AlCl3, what mass m of AlCl3 forms from 25 g of Al? Assume the other reactants are in excess.

M(Al) = 26.98 g/molM(AlCl3) = 133.33 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 12 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(NH3​)m(NH3​)÷ M(X)× 2/ν× M(NH3​)

24 g of N2 and 64 g of H2 react according to N2 + 3H2 → 2NH3. What mass m of NH3 can form?

M(N2) = 28.02M(H2) = 2.016M(NH3) = 17.03 g/mol
m = g
4.Medium
m(X)n(X)n(CO2​)m(CO2​)÷ M(X)× 3/ν× M(CO2​)

12 g of Fe2O3 and 20 g of CO react according to Fe2O3 + 3CO → 2Fe + 3CO2. What mass m of CO2 can form?

M(Fe2O3) = 159.69M(CO) = 28.01M(CO2) = 44.01 g/mol
m = g
5.Hard
m(C6​H12​O6​)n(C6​H12​O6​)n(C2​H5​OH)m(C2​H5​OH)÷ M(C6​H12​O6​)× 2/1× M(C2​H5​OH)

In the reaction C6H12O6 → 2C2H5OH + 2CO2, 20 g of C6H12O6 reacts completely (the other reactants are in excess) and 7.98 g of C2H5OH is collected. Find the percent yield Y.

M(C6H12O6) = 180.16 g/molM(C2H5OH) = 46.07 g/molactual: 7.98 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, 8 g of CH4 reacts completely (the other reactants are in excess) and 16.9 g of H2O is collected. Find the percent yield Y.

M(CH4) = 16.04 g/molM(H2O) = 18.02 g/molactual: 16.9 g
Y = %

Answer key

  1. 1. m = 123.5 g
  2. 2. m = 107.3 g
  3. 3. m = 29.17 g
  4. 4. m = 9.921 g
  5. 5. Y = 78.02 %
  6. 6. Y = 94.02 %

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