Question

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

Answer 1

Answer:

0.56 liters

Explanation:

At STP, one mole of a gaseous substance occupy 22.4 liters.

Hence 1 mole of oxygen gas will occupy 22.4 liters at STP.

We know that molar mass of oxygen is 32 g/mol.

So 32 grams of oxygen gas will occupy 22.4 liters at STP.

Then 0.8 grams of oxygen gas will occupy:

$\frac{22.4}{32}\times0.8=0.56$ liters.

Hence the volume of 0.8 grams of oxygen gas at STP is 0.56 liters.

Answer 2

Answer:

The volume of 0.80 grams of O2 is 0.56 L

Explanation:

Step 1 : Data given

Mass of O2 = 0.80 grams

Molar mass of O2 = 32 g/mol

STP = 1 mol, 1atm, 22.4 L

Step 2: Calculate moles of oxygen

Moles of O2 = Mass of O2 / molar mass of O2

Moles O2 = 0.80 grams / 32 g/mol

Moles O2 = 0.025 moles

Step 3: Calculate volume of O2

1 mol = 22.4 L of gas

0.025 moles = 0.025*22.4L = 0.56 L

The volume of 0.80 grams of O2 is 0.56 L