Question

What are the mole fractions of each component in a mixture of 15.08 g of O2, 8.17 g of N2, and 2.64 g of H2? What is the partial pressure in atm of each component of this mix ture if it is held in a 15.50-L vessel at 15°C?

Answer 1

Mole fraction = moles of one substance / total moles

15.08g O2 x (1 mol O2 / 32.0g O2) = 0.471 mol O2....X = 0.471 / 2.07 = 0.228

8.17g N2 x (1 mol N2 / 28.0g N2) = 0.292 mol N2 .... X = 0.292 / 2.07 = 0.141

2.64g H2 x (1 mol H2 / 2.02g H2) = 1.307 mol H2 .....X = 1.307 / 2.07 = 0.631

Total moles = 2.070 mol

Ptotal = nRT/V

Ptotal = 2.07 mol x 0.08206 Latm/molK x 288K / 15.50L = 3.156 atm

P(O2) = X(O2) P(total) = 0.228 x 3.156 atm = 0.720 atm

P(N2) = X(N2) P(total) = 0.141 x 3.156 atm = 0.445 atm

P(H2) = X(H2) P(total) = 0.631 x 3.156 atm = 1.992 atm