Question

16.

A mixture of methane and oxygen gases, at a total pressure of 701 mm Hg, contains 2.75 grams of methane and 3.45 grams of oxygen. What is the partial pressure of each gas in the mixture?

PCH4 = __
mm Hg
PO2 = ___
mm Hg

Answer 1

P(O₂) = 287.41 mmHg

P(O₂) = 413.59 mmHg

Step-by-step explanation:

Given data:

Total pressure = 701 mmHg

Mass of methane = 2.75 g

Mass of oxygen = 3.45 g

Partial pressure of each gas = ?

Solution:

Number of moles of methane:

Number of moles = mass/molar mass

Number of moles = 2.75 g/ 16 g/mol

Number of moles = 0.17 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 3.45 g/ 32 g/mol

Number of moles = 0.12 mol

Total number of moles = 0.12 mol + 0.17 mol = 0.29 mol

Partial pressure of oxygen:

P(O₂) = [ moles of oxygen / total moles ] × total pressure

P(O₂) = [0.12 / 0.29 ] × 701 mmHg

P(O₂) = 0.41 × 701 mmHg

P(O₂) = 287.41 mmHg

Partial pressure of methane:

P(O₂) = [ moles of oxygen / total moles ] × total pressure

P(O₂) = [0.17 / 0.29 ] × 701 mmHg

P(O₂) = 0.59 × 701 mmHg

P(O₂) = 413.59 mmHg