Question

If 237 g of H2 is mixed with N2 at a pressure of 152,000 torr and a temperature of 400°C, what volume of N2 gas is needed for the reaction?

Answer 1

n (H2) = m/M = 237 g / 2.02 g/mol = 117.33 mol

pV = nRT

V (H2) = nRT / P

T = 400 + 273 = 673 K

P = 152,000 torr = 20,265,000 Pa

V (H2) = 117.33 mol * 8.314 J/mol*K * 673 K / 20,265,000 Pa

V (H2) = 0.03 m³

V (N2) = 1/3 * 0.03 m³ = 0.01 m³ according to the equation:

N₂ + 3H₂ ⇄ 2NH₃