Question

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.
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Answer 1

Partial pressure of oxygen = 0.43 atm

Total pressure = 1.0 atm

Step-by-step 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