Question

A) Three gases (8.00 g of methane, CH4, 18.0 g of ethane, C2H6, and an unknown amount of propane, C3H8) were added to the same 10.0- L container. At 23.0 C, the total pressure in the container is 4.60atm . Calculate the partial pressure of each gas in the container.
Express the pressure values numerically in atmospheres, separated by commas. Enter the partial pressure of methane first, then ethane, then propane.
B) A gaseous mixture of O2 and N2 contains 39.8% nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 585mmHg ?

Answer 1

Answer:

A. PpCH4 = 1.21 atm, Pp C2H6 = 1.46 atm,  Pp C3H6 = 1.93 atm

B. Pp O2 = 352.17 mmHg

Explanation:

A. First step: Get the total mols of mixture to know the mass of propane.

P . V = n . R . T

T° K = T°C + 273 → 23°C + 273 = 296K

4.60 atm . 10L = n . 0.082 L.atm/mol.K . 296K

(4.60 atm . 10L) / (0.082 mol.K/L.atm . 296K) = n

1.89 mol = n

Mass/Molar weight = mol

Mass CH4 = 16 g/m

Mol CH4: 8g / 16g/m = 0.5 mol

Mass C2H6 = 30 g/m

Mol C2H6 = 0.6 mol

Total mols - Mol CH4 - Mol C2H6 = Mol propane

1.89 mols - 0.5 mol - 0.6 mol = 0.79 mol propane

2nd step: We can use the molar fraction with the partial pressure.

Partial pressure gas / Total pressure = mols gas / total moles

Sum of molar fraction = 1

0.5 / 1.89 + 0.6 /1.89 + 0.79/1.89 = 1

Partial pressure CH4 → Pp CH4 /4.60 atm =  0.5 / 1.89

PpCH4 = (0.5 /1.89) . 4.60 atm = 1.21 atm

Partial pressure C2H6 → Pp C2H6 /4.60atm = 0.6 /1.89

Pp C2H6 = (0.6 /1.89) . 4.60atm = 1.46 atm

Partial pressure C3H6 → Pp C3H6 /4.60atm = 0.79/1.89

Pp C3H6 = (0.79/1.89) . 4.60atm = 1.93 atm

B. 39.8% means a percent of the molar fraction, so the fraction is 0.398

N2 mass /total mass = Partial pressure N2/ Total pressure

0.398 . 585 mmHg = Partial pressure N2

232.83 mmHg = Partial pressure N2

Total pressure - Pp N2 = Pp O2

585 mmHg - 232.82 mmHg = 352.17 mmHg