Answer:
[tex]m=13.1gO_2[/tex]
Explanation:
Hello,
In this case, by using the ideal gas equation, we first compute the moles of oxygen at the given volume, pressure and temperature:
[tex]PV=nRT\\\\n=\frac{PV}{RT}=\frac{2.0atm*5L}{0.082\frac{atm*L}{mol*K}(25+273.15)K} =0.41mol[/tex]
Then, since molar mass of gaseous oxygen is 32 g/mol, we compute the contained mass in grams as shown below:
[tex]m=0.41mol*\frac{32g}{1mol}\\\\m=13.1gO_2[/tex]
Best regards.
