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(CH4​)n(CH4​)n(CO2​)m(CO2​)÷ M(CH4​)× 1/1× M(CO2​)

For the reaction CH4 + 2O2 → CO2 + 2H2O, what mass m of CO2 forms from 10 g of CH4? Assume the other reactants are in excess.

M(CH4) = 16.04 g/molM(CO2) = 44.01 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 8 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(H2​O)m(H2​O)÷ M(X)× 2/ν× M(H2​O)

4 g of H2 and 10 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(CO2​)m(CO2​)÷ M(X)× 1/ν× M(CO2​)

24 g of CH4 and 40 g of O2 react according to CH4 + 2O2 → CO2 + 2H2O. What mass m of CO2 can form?

M(CH4) = 16.04M(O2) = 32M(CO2) = 44.01 g/mol
m = g
5.Hard
m(C3​H8​)n(C3​H8​)n(H2​O)m(H2​O)÷ M(C3​H8​)× 4/1× M(H2​O)

In the reaction C3H8 + 5O2 → 3CO2 + 4H2O, 12 g of C3H8 reacts completely (the other reactants are in excess) and 13.73 g of H2O is collected. Find the percent yield Y.

M(C3H8) = 44.1 g/molM(H2O) = 18.02 g/molactual: 13.73 g
Y = %
6.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, 30 g of C6H12O6 reacts completely (the other reactants are in excess) and 8.44 g of C2H5OH is collected. Find the percent yield Y.

M(C6H12O6) = 180.16 g/molM(C2H5OH) = 46.07 g/molactual: 8.44 g
Y = %

Answer key

  1. 1. m = 27.44 g
  2. 2. m = 71.51 g
  3. 3. m = 11.26 g
  4. 4. m = 27.51 g
  5. 5. Y = 70 %
  6. 6. Y = 55.01 %

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