Question

Consider the balanced reaction below:

4Al+302=2Al203

How many grams of oxygen is needed to completely react with 9.30 moles of aluminum?

[ ] g O2

Answer 1

The mass needed is [223] g O₂.

M_r: 26.98 32.00

4Al + 3O₂ ⟶ 2Al₂O₃

n/mol: 9.30

Moles of O₂ = 9.30 mol Al × (3 mol O₂/4 mol Al) = 6.975 mol O₂

Mass of O₂= 6.975 mol O₂× (32.00 g O₂/1 mol O₂) = 223 g O₂