Question

A room with dimensions 7.00m×8.00m×2.50m is to be filled with pure oxygen at 22.0 °C and 1.00 atm. The molar mass of oxygen is 32.0 g/mol.

How many moles n oxygen of oxygen are required to fill the room?
What is the mass m oxygen of this oxygen?

Answer 1

1. 5765 mol

First of all, let's calculate the volume of the room (which corresponds to the volume of the gas):

$V=7.00 m\cdot 8.00 m \cdot 2.50 m=140 m^3$

We also know the following data about the gas:

$T=22.0^\circ +273 =295 K$ is the temperature

$p=1.00atm = 1.01\cdot 10^5 Pa$ is the pressure

Then we can use the ideal gas law

$pV=nRT$

with R being the gas constant

to find the number of moles of the gas:

$n=\frac{pV}{RT}=\frac{(1.01\cdot 10^5 Pa)(140 m^3)}{(8.314 J/mol K)(295 K)}=5765 mol$

2. 184.5 kg

The molar mass of oxygen is

$M_m = 32.0 g/mol$

this corresponds to the mass of 1 mol of oxygen.

In this problem, the number of moles is

n = 5765 mol

So the total mass of these n moles of oxygen will be:

$m=n M_m = (5765 mol)(32.0 g/mol)=1.845\cdot 10^5 g=184.5 kg$