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

a)

From the ideal gas equation,

PV=nRT

n=PV/RT

Here,
Volume,V= 7 m*8 m*2.5 m

                =140 m³

molar gas constant,R= 0.000082 m² g s^-2 K^-1 mol^-1
T= 22+273= 295 K
P= 1 atm
substitute the values in above expression,

n=(1 atm*140 m³)/(0.000082 m² g s^-2 K^-1 mol^-1*295 K)
=5788 moles

b)

1 mole of oxygen weight is 32 g
Therefore 5833moles of weight is ,

m=5788 moles*32 g

     = 185200.496 g