What is the mole fraction of O2 in a mixture of 5.06g of O2, 7.19g of N2, and 1.

ID: 874514 • Letter: W

Question

What is the mole fraction of O2 in a mixture of 5.06g of O2, 7.19g of N2, and 1.31g of H2?

What is the mole fraction of N2 in a mixture of 5.06g of O2, 7.19g of N2, and 1.31g of H2?

What is the mole fraction of H2 in a mixture of 5.06g of O2, 7.19g of N2, and 1.31g of H2?

What is the partial pressure in atm of O2 of this mixture if it is held in a 12.60?L vessel at 14?C?

What is the partial pressure in atm of N2 of this mixture if it is held in a 12.60?L vessel at 14?C?

What is the partial pressure in atm of H2 of this mixture if it is held in a 12.60?L vessel at 14?C?

What is the mole fraction of O2 in a mixture of 5.06g of O2, 7.19g of N2,

Mol O2 = Mass in g / Molar mass of O2 = 5.06 g O2 / 31.998 g/mol = 0.15813 mol O2

Mol N2 = 7.19 g / 28.014 g per mol = 0.257 mol N2

Mol H2 = 1.31 g / 2.0158 g per mol = 0.6498 mol H2

Mol fraction of O2 = mol O2 / Total mole in the mixture

= 0.15813 mol O2 / (0.257 +0.6498 +0.15813)

= 0.1485

Mol fraction of N2 = 0.257 mol N2 / 1.064 mol = 0.241

Mol fraction of H2 = 0.6498 mol H2/ 1.064 mol = 0.610

What is the partial pressure in atm of O2 of this mixture if it is held in a 12.60?L vessel at 4?C?

We use ideal gas law to find partial pressure of O2

pV= nRT

here p , V are pressure and volume. n = #mol , R = gas constant = 0.08206 L atm per K per mol , T = temperature in K

Lets plug given value of O2 in ideal gas law

P = nRT/ V

=[0.15813 mol O2 * (0.08206 Latm/Kmol)*277.15 K ] /12.60 L

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