What is the volume of 1.8 mol of O2 at STP

Answer 1

The easiest way to answer this would be to apply our understanding that 1 mole of any gas occupies 22.4 L of space at STP. To determine the volume, multiply the number of moles by this constant.

1.8 mol O₂ x 22.41 L/mol = 40.34 L O₂

Another way to approach answering this would be to use the ideal gas law, PV=nRT. The known values in this scenario include:

P = 1.0 atm (standard)

T = 273 K (standard)

R = 0.0821 (gas constant)

n = 1.8 mol

V = ? L

V = (1.8)(0.0821)(273) / (1.0) = 40.34 L O₂

Remember, if following significant figure conventions, the answer here would technically be 40. L O₂ because the provided 1.8 mol O₂ only contains two significant figures.

Answer 2

Hi there!

So, let's go step-by-step:

Molar mass of oxygen = 32 g/mol

Molar volume at STP (Standard temperature and pressure) = 22.4 L

If 32 g oxygen occupies 22.4 L at STP

Then,

1.8 mol oxygen occupies (22.4/32) × 1.8 = 1.26 L

Let me know if you have any questions!