The unbalanced reaction in question is
Al + O₂ → Al₂O₃
Let's balance it; let a, b, c be the coefficients in the reaction:
a Al + b O₂ → c Al₂O₃
Count up the atoms on either side for each element:
Al: a = 2c
O: 2b = 3c
Suppose c = 1; then
2b = 3c = 3 ⇒ b = 3/2
a = 2c = 2 ⇒ a = 2
So the balanced reaction is
2 Al + 3/2 O₂ → 1 Al₂O₃
or, using integer coefficients,
4 Al + 3 O₂ → 2 Al₂O₃
Now, we start with 5.4 kg of Al, which has molar mass of about 26.9815 g/mol. Convert the starting mass to moles:
(5.4 kg) (1000 g/kg) (1/26.9815 mol/g) ≈ 200.137 mol
In the balanced reaction, we have half as much Al₂O₃ produced for a given quantity of Al, so we end up with about 100.069 mol Al₂O₃ (molar mass ≈ 101.96 g/mol). Convert this amount to a mass:
(100.069 mol) (101.96 g/mol) (1/1000 kg/g) ≈ 10.203 kg ≈ 10. kg
