For the reaction 2Al + 3Cl2 → 2AlCl3, what mass m of AlCl3 forms from 64 g of Al? Assume the other reactants are in excess.
Stoichiometry worksheet with answers
1.Easy
M(Al) = 26.98 g/molM(AlCl3) = 133.33 g/mol
m = g
2.Easy
For the reaction 2Na + Cl2 → 2NaCl, what mass m of NaCl forms from 5 g of Na? Assume the other reactants are in excess.
M(Na) = 22.99 g/molM(NaCl) = 58.44 g/mol
m = g
3.Medium
24 g of Na and 32 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
4.Medium
6 g of Al and 80 g of Cl2 react according to 2Al + 3Cl2 → 2AlCl3. What mass m of AlCl3 can form?
M(Al) = 26.98M(Cl2) = 70.9M(AlCl3) = 133.33 g/mol
m = g
5.Hard
In the reaction 2Mg + O2 → 2MgO, 20 g of Mg reacts completely (the other reactants are in excess) and 31.17 g of MgO is collected. Find the percent yield Y.
M(Mg) = 24.31 g/molM(MgO) = 40.3 g/molactual: 31.17 g
Y = %
6.Hard
In the reaction 2Mg + O2 → 2MgO, 8 g of Mg reacts completely (the other reactants are in excess) and 11.67 g of MgO is collected. Find the percent yield Y.
M(Mg) = 24.31 g/molM(MgO) = 40.3 g/molactual: 11.67 g
Y = %
Answer key
- 1. m = 316.3 g
- 2. m = 12.71 g
- 3. m = 52.75 g
- 4. m = 29.65 g
- 5. Y = 94.01 %
- 6. Y = 88 %
Full step-by-step solutions: practise this topic on examproblems.com.