1) 5765 mol
First of all, we need to find the volume of the gas, which corresponds to the volume of the room:
[tex]V=7.00 m\cdot 8.00 m\cdot 2.50 m=140 m^3[/tex]
Now we can fidn the number of moles of the gas by using the ideal gas equation:
[tex]pV=nRT[/tex]
where
[tex]p=1.00 atm=1.01\cdot 10^5 Pa[/tex] is the gas pressure
[tex]V=140 m^3[/tex] is the gas volume
n is the number of moles
R is the gas constant
[tex]T=22.0^{\circ}+273=295 K[/tex] is the gas temperature
Solving for n,
[tex]n=\frac{pV}{RT}=\frac{(1.01\cdot 10^5 Pa)(140 m^3)}{(8.314 J/molK)(295 K)}=5765 mol[/tex]
2) 184 kg
The mass of one mole is equal to the molar mass of the oxygen:
[tex]M_m = 32.0 g/mol[/tex]
so if we have n moles, the mass of the n moles will be given by
[tex]m : n = 32.0 g/mol : 1 mol[/tex]
since n = 5765 mol, we find
[tex]m=\frac{(5765 mol)(32.0 g/mol)}{1 mol}=1.84\cdot 10^5 g=184 kg[/tex]
