Answer to Question #104551 in General Chemistry for Sofia

Question

a gaseous mixture (containing nitrogen, oxygen, and hydrogen) at 58 degrees celsius contains 2.42moles of nitrogen and 0.839 moles of oxygen in a 74.9L tank. If the total pressure is 992mmHG, what is the partial pressure of hydrogen gas? How many grams of hydrogen gas are contained in the mixture?

Answer 1

p * V = n * R * T

760 mm Hg - 1 atm

992 mm Hg - x atm

x = 1.3 atm

58°C = (58 + 273) K = 331 K

1.3 atm * 74.9 L = n * 0.0821 L*atm/(mol*K) * 331 K

n(mixture) = 3.58 mol

n(H₂) = n(mixture) - n(N₂) - n(O₂) = 3.58 mol - 2.42 mol - 0.839 mol = 0.321 mol

m(H₂) = n(H₂) * M(H₂) = 0.321 mol * 2 g/mol = 0.642 g

p(H₂) = n(H₂) / n(mixture) * p(mixture) = 0.321 mol / 3.58 mol * 1.3 atm = 0.117 atm = 89 mm Hg