For the reaction C6H12O6 → 2C2H5OH + 2CO2, what mass m of CO2 forms from 50 g of C6H12O6? Assume the other reactants are in excess.
Stoichiometry worksheet with answers
1.Easy
M(C6H12O6) = 180.16 g/molM(CO2) = 44.01 g/mol
m = g
2.Easy
For the reaction Fe2O3 + 3CO → 2Fe + 3CO2, what mass m of CO2 forms from 32 g of Fe2O3? Assume the other reactants are in excess.
M(Fe2O3) = 159.69 g/molM(CO2) = 44.01 g/mol
m = g
3.Medium
4 g of Fe2O3 and 48 g of CO react according to Fe2O3 + 3CO → 2Fe + 3CO2. What mass m of Fe can form?
M(Fe2O3) = 159.69M(CO) = 28.01M(Fe) = 55.85 g/mol
m = g
4.Medium
24 g of C3H8 and 20 g of O2 react according to C3H8 + 5O2 → 3CO2 + 4H2O. What mass m of CO2 can form?
M(C3H8) = 44.1M(O2) = 32M(CO2) = 44.01 g/mol
m = g
5.Hard
In the reaction 2Na + Cl2 → 2NaCl, 24 g of Na reacts completely (the other reactants are in excess) and 51.86 g of NaCl is collected. Find the percent yield Y.
M(Na) = 22.99 g/molM(NaCl) = 58.44 g/molactual: 51.86 g
Y = %
6.Hard
In the reaction Fe2O3 + 3CO → 2Fe + 3CO2, 4 g of Fe2O3 reacts completely (the other reactants are in excess) and 1.96 g of Fe is collected. Find the percent yield Y.
M(Fe2O3) = 159.69 g/molM(Fe) = 55.85 g/molactual: 1.96 g
Y = %
Answer key
- 1. m = 24.43 g
- 2. m = 26.46 g
- 3. m = 2.798 g
- 4. m = 16.5 g
- 5. Y = 85.01 %
- 6. Y = 70.05 %
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