Question

Part A: Three gases (8.00 g of methane, CH_4, 18.0g of ethane, C_2H_6, and an unknown amount of propane, C_3H_8) were added to the same 10.0- L container. At 23.0 degrees C, the total pressure in the container is 5.40 atm. Calculate the partial pressure of each gas in the container.Part B: A gaseous mixture of O_2 and N_2 contains 37.8 % nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 405 mmHg?

Answer 1

Step-by-step explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

`n=(PV)/(RT)=(5.40 atm* 10.0 L)/(0.0821 atm L/mol K* 296.15 K)=2.22 mol`

Moles of methane gas =
`n_1=(8.00 g)/(16 g/mol)=0.5 mol`

Moles of ethane gas =
`n_2=(18.0 g)/(30 g/mol)=0.6 mol`

Moles of propane gas =
`n_3`

`n=n_1+n_2+n_3`

`2.22=0.5 mol +0.6 mol+ n_3`

`n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol`

Mole fraction of methane =
`\chi_1=(n_1)/(n_1+n_2+n_3)=(n_1)/(n)`

`\chi_1=(0.5 mol)/(2.22 mol)=0.2252`

Similarly, mole fraction of ethane and propane :

`\chi_2=(n_2)/(n)=(0.6 mol)/(2.22 mol)=0.2703`

`\chi_3=(n_3)/(n)=(1.12 mol)/(2.22 mol)=0.5045`

Partial pressure of each gas can be calculated by the help of Dalton's' law:

`p_i=P* \chi_1`

Partial pressure of methane gas:

`p_1=P* \chi_1=5.40 atm* 0.2252=1.22 atm`

Partial pressure of ethane gas:

`p_2=P* \chi_2=5.40 atm* 0.2703=1.46 atm`

Partial pressure of propane gas:

`p_3=P* \chi_3=5.40 atm* 0.5045=2.72 atm`

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas =
`n_1=(37.8 g g)/(28g/mol)=1.35 mol`

Moles of oxygen gas =
`n_2=(62.2 g)/(32 g/mol)=1.94 mol`

Mole fraction of nitrogen=
`\chi_1=(n_1)/(n_1+n_2)`

`\chi_1=(1.35 mol)/(1.35 mol+1.94 mol)=0.4103`

Similarly, mole fraction of oxygen

`\chi_2=(n_2)/(n_1+n_2)=(1.94 mol)/(1.35 mol+1.94 mol)=0.5897`

Partial pressure of each gas can be calculated by the help of Dalton's' law:

`p_i=P* \chi_1`

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

`p_1=P* \chi_1=405 mmHg* 0.4103 =166.17 mmHg`

Partial pressure of oxygen gas:

`p_2=P* \chi_2=405 mmHg* 0.5897=238.83 mmHg`