Question

Find the mass of Al2O3 that will be formed when 9g of Al burns in 16g of O2.​

Answer 1

Answer: 4Al +2 O2 =2Al2O3

Explanation:

9g Al =9/27=0.33 mole

16/32= 0.5mole O2

Limiting reagent is Al

So product formation depends on Al

From eqn 4 moles of Al gives 2 moles of Al2O3

0.33 moles of. Al give...?

0.33x2/4 =0.165 moles of Al2O3

Weight of al2o3 =0.165x102=16.83 gm