Question

How many grams of O2 are present in 44.1 L of O2 at STP?

Answer 1

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:

• P is the gas pressure.
• V is the volume that occupies.
• T is its temperature.
• R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
• n is the number of moles of the gas.

Then, in this case:

• P= 1 atm
• V= 44.1 L
• n= ?
• R= 0.082 $\frac{atmL}{molK}$
• T= 0°C =273 K

Replacing in the expression for the ideal gas law:

1 atm× 44.1 L= n× 0.082 $\frac{atmL}{molK}$× 273 K

Solving:

$n=\frac{1 atm x44.1 L}{0.082\frac{atmL}{molK}x273K}$

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:

$1.97 molesx\frac{32 g}{1 mole}$= 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:

• brainly.com/question/4147359?referrer=searchResults