Question

Calculate the partial pressure of each in atm for a gas mixture that contains 1.077 grams of oxygen gas, 1.215 grams of nitrogen gas, and 3.350 grams of carbon dioxide gas in a 5.00 L vessel at 700mmHg, and 37.0 degrees Celsius.

Oxygen =

Nitrogen =

Carbon dioxide =

Answer 1

mass of oxygen = 1.077 gm, moles of oxygen = mass of oxygen/ molar mass, molar mass of oxygen =32

moles of oxygen = 1.077/32=0.034, moles of ntrogen = mass of nitrogen/molar mass , molar mass of nitrogen= 28

moles of nitrogen= 1.215/28=0.0434, moles of CO2= mass of CO2/ molar mass, molar mass of CO2=44

moles of CO2= 3.35/44 =0.076

total moles of mixture= moles of Oxygen+ moles of nitrogen+ moles of CO2= 0.034+0.0434+0.076=0.1534

moles fraction= moles/total moles

mole fraction: Oxygen =0.034/0.1534 =0.22, N2= 0.0434/0.1534= 0.28, CO2= 0.50

total pressure = 700mm Hg, 760 mm Hg= 1 atm, 700mm Hg= 700/760 atm =0.92 atm

partial pressure : mole fraction* total pressure

partial pressure (atm) : O2= 0.22*0.92=0.20, O2=0.28*0.92= 0.26, CO2= 0.5*0.92=0.46