A gas mixture containing 2.6 moles of H2 and 4.3 moles of He is at a total pressure of 100 kPa. What is the partial pressure of helium in the mixture?



Answer :

Answer:

The partial pressure of He is 62.32 kPa.

Explanation:

The given information from the exercise is:

- Moles of H2 (nH2): 2.6 moles

- Moles of He (nHe): 4.3 moles

- Total pressure (PT): 100kPa

1st) To calculate the partial pressure, it is necessary to know the total number of moles (nT), so we have to add moles of H2 plus moles of He:

[tex]\begin{gathered} n_T=n_{H_2}+n_{He} \\ n_T=2.6\text{ moles}+4.3\text{ moles} \\ n_T=6.9\text{ moles} \end{gathered}[/tex]

The total number of moles (nT) is 6.9.

2nd) Now we can use the partial pressure formula, to calculate the partial pressure of helium (He), by replacing the values of nHe, nT and PT:

[tex]\begin{gathered} P_{He}=\frac{n_{He}}{n_T}*P_T \\ P_{He}=\frac{4.3\text{ moles}}{6.9\text{ moles}}*100kPa \\ P_{He}=62.32kPa \end{gathered}[/tex]

Finally, the partial pressure of He is 62.32 kPa.

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