Limiting reactant : Al
Al2O3 produced = 18.36 g
mass O2 remains = 11.904 g
Further explanation
Given
Reaction
4Al + 3O2 —> 2Al2O3
10.7 g O2
Required
Limiting reactant
mass Al2O3
Mass of excess reactant remains
Solution
mol Al : 10.7 : 27 = 0.396
mol O2 = 10.7 : 32 = 0.334
1. limiting reactant
mol : coefficient of Al : O2 = 0.396/4 : 0.334/3 = 0.099 : 0.111
Al as a limitng reactant(smaller ratio)
2. mass Al2O3
mol Al2O3 based on Al : 2/4 x 0.396 = 0.18
mass Al2O3(MW=102 g/mol) = 0.18 x 102 = 18.36 g
3. mass O2 remains
mol O2 reacted : 3/4 x 0.396 = 0.297
mol O2 remains = 0.669 - 0.297 =0.372
mass O2 remains = 0.372 x 32 = 11.904 g