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How many grams of O2 are present in 44.1 L of O2 at STP?
a.10.0 g
b.16.0 g
c.32.0 g
d.410.0 g
e. 63.0 g

Sagot :

CintiaSalazaridn

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 [tex]\frac{atmL}{molK}[/tex]
  • T= 0°C =273 K

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:

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