Answer:
[tex]m_{air}=0.947g[/tex]
[tex]n_{O_2} =0.00686molO_2[/tex]
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:
[tex]T=212\°F=100\°C=373.15K\\\\n=\frac{PV}{RT}=\frac{1.00atm*1.00L}{0.082\frac{atm*L}{mol*K}*373.15K}\\ \\n=0.0327mol[/tex]
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:
[tex]m_{air}=0.0327mol*\frac{28.97g}{1mol} =0.947g[/tex]
Finally, knowing that the 21% of the 0.0327 moles of air is oxygen, its moles turn out:
[tex]n_{O_2}=0.0327mol*\frac{0.21molO_2}{1mol} =0.00686molO_2[/tex]
Best regards.
