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Question: An oxygen gas container has a volume of \( 20.0\text{ }L.~ \) How many grams of oxygen are in the co...

An oxygen gas container has a volume of 20.0 L. 20.0\text{ }L.~ How many grams of oxygen are in the container if the gas has a pressure of 876mmHg876mmHg at 23oC{{23}^{o}}C ?

Explanation

Solution

Hint : The ideal gas law (PV=nRT)\left( PV=nRT \right) can be used to find the value of total pressure of the container. The number of moles of both hydrogen and oxygen need to be found and added to get the total number of moles in the container. Formula used: The number of moles can be found using the equation:
Number(mole)=GivenMassMolarMass.Number(mole)=\dfrac{GivenMass}{MolarMass}.

Complete Step By Step Answer:
In order to calculate the total pressure of the container, we can use the ideal gas law (PV=nRT)\left( PV=nRT \right) we have to find the total pressure (P)(P) of the container. Hence, it is necessary that we find the total number of moles in the container. So, we have to find the number of moles of both hydrogen and oxygen and then add them to get the total number of moles in the container. For calculating the number of moles of hydrogen and oxygen, we can use the equation: Number(mole)=GivenMassMolarMass.Number(mole)=\dfrac{GivenMass}{MolarMass}.
Use the ideal gas law. We are given the pressure, the temperature and the volume. The gas constant will be 62.36,62.36, (PV=nRT)\left( PV=nRT \right) we're looking for n, the number of moles in the sample. Note that;
23oC=23+273=296K.{{23}^{o}}C=23+273=296K.
Now we have, (20L)(86mmHg)=n(62.36mmHgL⋅k−1mol−1)(296K)(20L)(86mmHg)=n(62.36mmHgL\cdot {{k}^{-1}}mo{{l}^{-1}})\left( 296K \right)
⇒n=0.949 mol\Rightarrow n=0.949~mol
We know that Number(mole)=GivenMassMolarMass.Number(mole)=\dfrac{GivenMass}{MolarMass}.
It can be re-written as; MolarMass=GivenMassNumber(mole).MolarMass=\dfrac{GivenMass}{Number(mole)}.
And that the molar mass of oxygen is 32.00g⋅mol−132.00g\cdot mo{{l}^{-1}} we can solve for the number of grams in the sample.
⇒g=(32g⋅mol−1)×(0.949mol)\Rightarrow g=\left( 32g\cdot mo{{l}^{-1}} \right)\times \left( 0.949mol \right)
⇒g=30.4grams\Rightarrow g=30.4grams
Therefore, 30.430.4 grams of oxygen are in the container if the gas has a pressure of 876mmHg876mmHg at 23oC.{{23}^{o}}C.

Note :
Note that it is important that you convert the temperature from degree Celsius to kelvin. This is because the Kelvin scale is the SI unit of temperature. Using the value of zero degrees Celsius will give you the value of pressure to be zero, which is incorrect.