Question

1. What is the mole fraction of O2 in a mixture of 5.07 g of O2, 7.16 g of N2, and 1.33 g of H2?

2. What is the mole fraction of N2 in a mixture of 5.07 g of O2, 7.16 g of N2, and 1.33 g of H2?

3. What is the mole fraction of H2 in a mixture of 5.07 g of O2, 7.16 g of N2, and 1.33 g of H2?

4. What is the partial pressure in atm of O2 of this mixture if it is held in a 12.50 −L vessel at 13 ∘C?

5. What is the partial pressure in atm of N2 of this mixture if it is held in a 12.50 −L vessel at 13 ∘C?

6. What is the partial pressure in atm of H2 of this mixture if it is held in a 12.50 −L vessel at 13 ∘C?

Answer 1

Answer - We are given, 5.07 g of O₂, 7.16 g of N₂, and 1.33 g of H₂

First we need to calculate the moles of each

Moles of O₂ = 5.07 g / 31.998 g.mol^-1 = 0.158 moles

Moles of N₂ = 7.16 g / 28.014 g.mol^-1 = 0.256 moles

Moles of H₂ = 1.33 g / 2.016 g.mol^-1 = 0.660 moles

Now we need to calculate the total moles

Total moles = 0.158 + 0.256 + 0.660

                     = 1.07 moles

We know mole fraction formula

Moles fraction = moles / total moles

1) Moles fraction of O₂ = 0.158 moles / 1.07 mole = 0.148

2) Moles fraction of N₂ = 0.256 moles / 1.07 mole = 0.238

3) Moles fraction of H₂ = 0.660 moles / 1.07 mole = 0.614

Now we need to calculate the total pressure from the total moles, volume and temperature given

Total moles = 1.07 moles , V = 12.50 L , T = 13 +273 = 286 K

Now using the Ideal gas law –

PV = nRT

P = nRT /V

   = 1.07 moles * 0.0821 L.atm.mol^-1.K^-1*286 K / 12.50 L

   = 2.02 atm

We know the formula

Mole fraction = partial pressure / total pressure

So, partial pressure = mole fraction * total pressure

4 ) Partial pressure of O₂ = 0.148 * 2.02 atm = 0.298 atm

5) Partial pressure of N₂ = 0.238 * 2.02 atm = 0.480 atm

6) Partial pressure of H₂ = 0.614 * 2.02 atm = 1.24 atm