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(C2​H5​OH)m(C2​H5​OH)÷ M(C6​H12​O6​)× 2/1× M(C2​H5​OH)

For the reaction C6H12O6 → 2C2H5OH + 2CO2, what mass m of C2H5OH forms from 64 g of C6H12O6? Assume the other reactants are in excess.

M(C6H12O6) = 180.16 g/molM(C2H5OH) = 46.07 g/mol
m = g
2.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 5 g of CH4? Assume the other reactants are in excess.

M(CH4) = 16.04 g/molM(CO2) = 44.01 g/mol
m = g
3.Medium
m(X)n(X)n(NH3​)m(NH3​)÷ M(X)× 2/ν× M(NH3​)

12 g of N2 and 8 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(NaCl)m(NaCl)÷ M(X)× 2/ν× M(NaCl)

10 g of Na and 20 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(Al)n(Al)n(AlCl3​)m(AlCl3​)÷ M(Al)× 2/2× M(AlCl3​)

In the reaction 2Al + 3Cl2 → 2AlCl3, 16 g of Al reacts completely (the other reactants are in excess) and 49.02 g of AlCl3 is collected. Find the percent yield Y.

M(Al) = 26.98 g/molM(AlCl3) = 133.33 g/molactual: 49.02 g
Y = %
6.Hard
m(CaCO3​)n(CaCO3​)n(CO2​)m(CO2​)÷ M(CaCO3​)× 1/1× M(CO2​)

In the reaction CaCO3 → CaO + CO2, 16 g of CaCO3 reacts completely (the other reactants are in excess) and 5.28 g of CO2 is collected. Find the percent yield Y.

M(CaCO3) = 100.09 g/molM(CO2) = 44.01 g/molactual: 5.28 g
Y = %

Answer key

  1. 1. m = 32.73 g
  2. 2. m = 13.72 g
  3. 3. m = 14.59 g
  4. 4. m = 25.42 g
  5. 5. Y = 62 %
  6. 6. Y = 75.05 %

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