Question

How much space would .33 moles of oxygen take up at STP?​

Answer 1

Answer:

0.33 moles of Oxygen at stp would occupy a volume of 7.392 dm³

Explanation:

1 mole of every gas at standard temperature and pressure (stp) occupies 22.4 dm³

0.33 moles of Oxygen at stp would occupy a volume of 0.33 × 22.4 dm³ = 7.392 dm³

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Answer 2

Answer:

$V = 7.391\,L$

Explanation:

Let suppose that oxygen behaves ideally. The equation of state of the ideal gas is:

$P\cdot V = n\cdot R_{u}\cdot T$

The volume is now cleared:

$V = \frac{n\cdot R_{u}\cdot T}{P}$

$V = \frac{(0.33\,mol)\cdot \left(0.082\,\frac{atm\cdot L}{mol\cdot K} \right)\cdot (273.15\,K)}{1\,atm}$

$V = 7.391\,L$