Taking into accoun the STP conditions and the ideal gas law, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.
First of all, the STP conditions refer to the standard temperature and pressure, where the values used are: pressure at 1 atmosphere and temperature at 0°C. These values are reference values for gases.
On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:
P×V = n×R×T
where:
Then, in this case:
Replacing in the expression for the ideal gas law:
1 atm× 44.1 L= n× 0.082 [tex]\frac{atmL}{molK}[/tex]× 273 K
Solving:
[tex]n=\frac{1 atm x44.1 L}{0.082\frac{atmL}{molK}x273K}[/tex]
n=1.97 moles
Being the molar mass of O₂, that is, the mass of one mole of the compound, 32 g/mole, the amount of mass that 1.97 moles contains can be calculated as:
[tex]1.97 molesx\frac{32 g}{1 mole}[/tex]= 63.04 g ≈ 63 g
Finally, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.
Learn more about the ideal gas law:
