Part A What is the mole fraction of O2 in a mixture of 15.1 g of O2, 8.18 g of N
ID: 973091 • Letter: P
Question
Part A What is the mole fraction of O2 in a mixture of 15.1 g of O2, 8.18 g of N2, and 2.46 g of H2? Part B What is the mole fraction of N2 in a mixture of 15.1 g of O2, 8.18 g of N2, and 2.46 g of H2? Part C What is the mole fraction of H2 in a mixture of 15.1 g of O2, 8.18 g of N2, and 2.46 g of H2? Part D What is the partial pressure in atm of O2 of this mixture if it is held in a 15.30 L vessel at 17 C? Part E What is the partial pressure in atm of N2 of this mixture if it is held in a 15.30 L vessel at 17 C? Part F What is the partial pressure in atm of H2 of this mixture if it is held in a 15.30 L vessel at 17 C?
Explanation / Answer
Parts A through C
Moles of O2 = 15.1g * (1 mol / 32 g) = 0.4719 moles
Moles of N2 = 8.18 g * (1mol / 28 g) = 0.2921 moles
Moles of H2 = 2.46 g * (1mol / 2g) = 1.23 moles
Mole fraction of O2 = 0.4719 / (0.4719 + 0.2921 + 1.23) = 0.2366
Mole fraction of N2 = 0.2921 / (0.4719 + 0.2921 + 1.23) = 0.1465
Mole fraction of H2 = 1.23 / (0.4719 + 0.2921 + 1.23) = 0.6168
Parts D through F
Total pressure = nRT / V = 1.994 * 0.082 * (17 + 273) / 15.3 L = 3.1 atm
Partial pressure of O2 = 0.2366 * 3.1 = 0.7333 atm
Partial pressure of N2 = 0.1465 * 3.1 = 0.45415 atm
Partial pressure of H2 = 0.6168 * 3.1 = 1.9121 atm