Question

Suppose you have a container filled with air at 212 oF. The volume of the container 1.00 L, the pressure of air is 1.00 atm. The molecular composition of air is 79% N2 and 21% O2 for simplification. Calculate the mass of air and moles of O2 in the container.

Answer 1

`m_(air)=0.947g`

`n_(O_2) =0.00686molO_2`

Step-by-step explanation:

Hello,

In this case, we can firstly use the ideal gas equation to compute the total moles of the gaseous mixture (air) with the temperature in Kelvins:

`T=212\°F=100\°C=373.15K\\\\n=(PV)/(RT)=(1.00atm*1.00L)/(0.082(atm*L)/(mol*K)*373.15K)\\ \\n=0.0327mol`

In such a way, since the molar mass of air is 28.97 g/mol, we can compute the mass of air with a single mass-mole relationship:

`m_(air)=0.0327mol*(28.97g)/(1mol) =0.947g`

Finally, knowing that the 21% of the 0.0327 moles of air is oxygen, its moles turn out:

`n_(O_2)=0.0327mol*(0.21molO_2)/(1mol) =0.00686molO_2`

Best regards.