Question

Calculate the number of grams of oxygen gas in a 24L container at STP

Answer 1

There are 34 g of oxygen in the container.

We can use the Ideal Gas Law to solve this problem.

$pV = nRT$

But $n = \frac{m}{M}$, so

$pV = \frac{m}{M}RT$ and

$m = \frac{pVM}{RT}$

STP is 0 °C and 1 bar, so

$m = \frac{\text{1 bar} \times \text{24 L} \times 32.00 \text{ g}\cdot\text{mol}^{-1}}{\text{0.083 14 } \text{bar}\cdot\text{L}\cdot\text{K}^{-1}\text{mol}^{-1}\times\text{273.15 K} } = \textbf{34 g}$