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
[tex]n = \text{17.4 mol Al}\times \dfrac{\text{3 mol O}_{2}}{\text{4 mol Al}}= \text{13.05 mol O}_{2}[/tex]
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.
[tex]V = \text{13.05 mol}\times \dfrac{\text{22.71 L}}{\text{1 mol }}= \textbf{296 L}[/tex]
