Question

A mixture of hydrogen and neon gases is maintained in a 5.36 L flask at a pressure of 2.65 atm and a temperature of 82 °C. If the gas mixture contains 0.706 grams of hydrogen, the number of grams of neon in the mixture is______g.

Answer 1

n(H₂) = 0.706 / 2 = 0.353 mol

V(mix) = 0.00536 m³ (mix mean mixture)

p(mix) = 2.65 * 101325 = 268500 Pa

T(mix) = 82 + 273 = 355 K

n(mix) = n(Ne) + n(H₂) = pV/(RT) = 268500*0.00536/(8.314*355) = 0.488 mol

n(Ne) = 0.488 - 0.353 = 0.135 mol

m(Ne) = 0.135 * 20.2 = 2.72 g

Answer: 2.72 g