Question

How many liters of oxygen are required to react completely with 14.8 mol of Al?

Answer 1

Answer:

296 L  

Explanation:

We will need a balanced equation with moles, so let's gather all the information in one place.

                  4Al + 3O₂ ⟶ 2Al₂O₃

n/mol:        17.4

1. Moles of O₂

$n = \text{17.4 mol Al}\times \dfrac{\text{3 mol O}_{2}}{\text{4 mol Al}}= \text{13.05 mol O}_{2}$

2. Volume of O₂

You haven't given the conditions at which the volume is measured, so I assume it is at STP (0 °C and 1 bar).

At STP the molar volume of a gas is 22.71 L.

$V = \text{13.05 mol}\times \dfrac{\text{22.71 L}}{\text{1 mol }}= \textbf{296 L}$