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
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
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