Question

A mixture of gases contains 1.26 g of N2, 3.71 g of H2, and 1.87 g of NH3. If the total pressure of the mixture is 1.67 atm, what is the partial pressure of each component?

PN2 = ? atm

PH2 = ? atm

PNH3 = ? atm

Answer 1

Answer

moles of N2 = mass/molar mass = 1.26 g/ 28 g/mol = 0.045 moles

moles of H2= mass/molar mass= 3.71 g/2 g/mol = 1.855 moles

moles of NH3 = mass/molar mass = 1.87 g/ 17.031 g/mol = 0.1097 moles

Partial pressure of N2 = mole fraction of N2*total pressure

= (moles of N2/(moles of N2+H2+NH3))*total pressure

  = (0.045/(0.045+1.855+0.1097))*1.67

= 0.037 atm

Partial pressure of H2 = mole fraction of H2*total pressure

= (moles of H2/(moles of N2+H2+NH3))*total pressure

= (1.855/(0.045+1.855+0.1097))*1.67

= 1.54 atm

Partial pressure of NH3 = mole fraction of NH3*total pressure

  = (moles of NH3/(moles of N2+H2+NH3))*total pressure

= (0.1097/(0.045+1.855+0.1097))*1.67

= 0.091 atm