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 noxygen of oxygen are required to fill the room?
What is the mass moxygen of this oxygen?

Answer 1

1) 5765 mol

First of all, we need to find the volume of the gas, which corresponds to the volume of the room:

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

Now we can fidn the number of moles of the gas by using the ideal gas equation:

$pV=nRT$

where

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

$V=140 m^3$ is the gas volume

n is the number of moles

R is the gas constant

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

Solving for n,

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

2) 184 kg

The mass of one mole is equal to the molar mass of the oxygen:

$M_m = 32.0 g/mol$

so if we have n moles, the mass of the n moles will be given by

$m : n = 32.0 g/mol : 1 mol$

since n = 5765 mol, we find

$m=\frac{(5765 mol)(32.0 g/mol)}{1 mol}=1.84\cdot 10^5 g=184 kg$