1 Answer

Aqfaridi · Answer 1 · 2021-12-17T22:31:03+0000

Answer:

296 L

Step-by-step 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

$n = \text{17.4 mol Al}* \frac{\text{3 mol O}_(2)}{\text{4 mol Al}}= \text{13.05 mol O}_(2)$

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.

$V = \text{13.05 mol}* \frac{\text{22.71 L}}{\text{1 mol }}= \textbf{296 L}$

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