Question

What is the volume of 0.80 grams of O2 gas at stp?

Answer 1

Answer : The volume of oxygen gas at STP is, 0.56 L or 560 ml

Explanation : Given,

Mass of oxygen gas = 0.80 g

Molar mass of oxygen gas = 32 g/mole

First we have to calculate the moles of oxygen gas.

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{0.80g}{32g/mole}=0.025mole$

Now we have to calculate the volume of oxygen gas.

As we know that at STP,

1 mole of oxygen gas contains 22.4 L volume if oxygen gas

So, 0.025 mole of oxygen gas contains $0.025\times 22.4L=0.56L=560ml$ volume if oxygen gas

Conversion : 1 L = 1000 ml

Therefore, the volume of oxygen gas at STP is, 0.56 L or 560 ml

Answer 2

Answer: The volume occupied by $O_2$ at STP is 0.56 L.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

We are given:

Given mass of oxygen = 0.80 g

Molar mass of oxygen = 32 g/mol

Putting values in above equation, we get:

$\text{Moles of oxygen}=\frac{0.80g}{32g/mol}=0.025mol$

At STP:

1 mole of a gas occupies 22.4 L of volume.

So, 0.025 moles of oxygen gas will occupy = $22.4\times 0.025=0.56L$ of volume.

Hence, the volume occupied by $O_2$ at STP is 0.56 L.