Question

A mixture of gases consists of 4.00 moles of He, 2.00 moles of H2, 3.00 moles of CO2 and 5.00 moles of Ar. The total pressure of the mixture is 2900 mm. Determine the mole fraction of each gas in the mixture. Determine the mole percent of each gas in the mixture. Determine the partial pressure of each gas in the mixture.

Answer 1

Step-by-step explanation:

Moles of helium gas =
`n_1=4.00 mol`

Moles of hydrogen gas =
`n_2=2.00 mol`

Moles of carbon dioxide gas =
`n_3=3.00 mol`

Moles of argon gas =
`n_4=5.00 mol`

Mole fraction of helium gas =
`\chi_1=(n_1)/(n_1+n_2+n_3+n_4)=(4.00 mol)/(14.00 mol)=0.2857`

Mole fraction of hydrogen gas =
`\chi_2=(n_2)/(n_1+n_2+n_3+n_4)=(2.00 mol)/(14.00 mol)=0.1428`

Mole fraction of carbon dioxide gas =
`\chi_3=(n_3)/(n_1+n_2+n_3+n_4)=(3.00 mol)/(14.00 mol)=0.2143`

Mole fraction of argon gas =
`\chi_4=(n_4)/(n_1+n_2+n_3+n_4)=(5.00 mol)/(14.00 mol)=0.3571`

Mole percent of helium =
`\chi_1* 100=0.2857* 100=28.57\%`

Mole percent of hydrogen=
`\chi_2* 100=0.1428 * 100=14.28\%`

Mole percent of helium =
`\chi_3* 100=0.2143* 100=21.43\%`

Mole percent of helium =
`\chi_4* 100=0.3571 * 100=35.71\%`

Total pressure of the mixture = P = 2900 mmHg

Partial pressure of helium gas =
`p_1`

Partial pressure of hydrogen gas =
`p_2`

Partial pressure of carbon dioxide gas =
`p_3`

Partial pressure of argon gas =
`p_4`

`p_i=P* \chi_i`

`chi_i` = Mole fraction of ith component

`p_1=P* \chi_1=P* (n_1)/(n_1+n_2+n_3+n_4)`

`p_1=2900 mmHg* (4.00 mol)/(4.00 mol+2.00 mol+3.00 mol+5.00 mol)=828.57 mmHg`

`p_2=P* \chi_2=P* (n_2)/(n_1+n_2+n_3+n_4)`

`p_2=2900 mmHg* (2.00 mol)/(4.00 mol+2.00 mol+3.00 mol+5.00 mol)=828.57 mmHg`

`p_3=P* \chi_3=P* (n_3)/(n_1+n_2+n_3+n_4)`

`p_3=2900 mmHg* (3.00 mol)/(4.00 mol+2.00 mol+3.00 mol+5.00 mol)=414.28 mmHg`

`p_4=P* \chi_4=P* (n_4)/(n_1+n_2+n_3+n_4)`

`p_4=2900 mmHg* (5.00mol)/(4.00 mol+2.00 mol+3.00 mol+5.00 mol)=1,035.71 mmHg`