Volume of Water Vapor Collected ?

Question

You collect 2.6 grams of hydrogen gas over water at a temperature of 75^OC and the partial pressure of hydrogen is 1.6 atm. What are the grams of the water vapor and the volume of the water vapor collected?

Answer 1

Pressure of water vapor at 75ºC = 0.38 atm (I looked it up)

Total pressure = pressure water vapor + pressure H₂ gas = 0.38 atm + 1.6 atm = 1.98 atm

moles H₂ = 2.6 g x 1 mol/2 g = 1.3 moles H₂

0.38 atm/1.98 atm = 0.1919 = mole fraction H₂O

let x = moles H₂O

x/1.3 + x = 0.1919

x = 0.308 moles H₂O

mass H₂O = 0.308 moles x 18 g/mol = 5.5 g H₂O (to 2 significant figures)

V = nRT/P = (0.308 mol)(0.0821 Latm/Kmol)(348K) / 0.38 atm

V = 23 L