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(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 50 g of Na? Assume the other reactants are in excess.

M(Na) = 22.99 g/molM(NaCl) = 58.44 g/mol
m = g
2.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 12 g of N2? Assume the other reactants are in excess.

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

12 g of H2 and 80 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(MgO)m(MgO)÷ M(X)× 2/ν× M(MgO)

30 g of Mg and 16 g of O2 react according to 2Mg + O2 → 2MgO. What mass m of MgO can form?

M(Mg) = 24.31M(O2) = 32M(MgO) = 40.3 g/mol
m = g
5.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, 8 g of C6H12O6 reacts completely (the other reactants are in excess) and 2.74 g of CO2 is collected. Find the percent yield Y.

M(C6H12O6) = 180.16 g/molM(CO2) = 44.01 g/molactual: 2.74 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, 12 g of CH4 reacts completely (the other reactants are in excess) and 22.11 g of H2O is collected. Find the percent yield Y.

M(CH4) = 16.04 g/molM(H2O) = 18.02 g/molactual: 22.11 g
Y = %

Answer key

  1. 1. m = 127.1 g
  2. 2. m = 14.59 g
  3. 3. m = 90.1 g
  4. 4. m = 40.3 g
  5. 5. Y = 70.1 %
  6. 6. Y = 82 %

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