15.A mixture of methane and oxygen gases, in a 8.77 L flask at 80 °C, contains 3.25 grams of methane and 4.08 grams of oxygen. The partial pressure of oxygen in the flask is atm and the total pressure in the flask is
atm.
Submit Answer

Partial pressure of oxygen =  0.43 atm

Total pressure = 1.0 atm

Explanation:

Given data:

Volume of flask = 8.77 L

Temperature = 80°C (80 + 273.15 K = 353.15 K)

Mass of methane = 3.25 g

Mass of oxygen = 4.08 g

Partial pressure of oxygen = ?

Total pressure = ?

Solution:

Number of moles of methane:

Number of moles = mass/molar mass

Number of moles = 3.25 g/ 16 g/mol

Number of moles = 0.2 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 4.08 g/ 32 g/mol

Number of moles = 0.13 mol

Total number of moles in flask = 0.13 mol + 0.2 mol = 0.33 mol

Pressure of oxygen:

PV = nRT

R = general gas constant = 0.0821 atm.L/mol.K

P ×8.77 L = 0.13 mol × 0.0821 atm.L/mol.K  × 353.15 K

P = 3.77atm.L /8.77 L

P =  0.43 atm

Pressure of flask:

PV = nRT

R = general gas constant = 0.0821 atm.L/mol.K

P ×8.77 L = 0.33 mol × 0.0821 atm.L/mol.K  × 353.15 K

P = 8.75atm.L /8.77 L

P =  1.0 atm

answer: mole of the gas

answer: mole of the gasexplanation: