Question

1.30 g each of CO, H2O, and H2 are sealed in a 1.35 −L vessel and brought to equilibrium at 600 K. CO(g)+H2O(g)⇌CO2(g)+H2(g)KC=23.2 A) How many grams of CO2 will be present in the equilibrium mixture?

Answer 1

we know that

moles = mass / molar mass

so

moles of CO = 1.3 / 28 = 0.04643

moles of H20 = 1.3 / 18 = 0.07222

moles of H2 = 1.3 / 2 = 0.65

now

concentration = moles / volume (L)

so

[CO] = 0.04643 / 1.35 = 0.03439

[H20] = 0.07222 / 1.35 = 0.0535

[H2] = 0.65 / 1.35 = 0.4815

now

consider the reaction

CO + H20 ---> C02 + H2

using ICE table

at equilibrium

[C0] = 0.03439 - x

[H20] = 0.0535 - x

[C02] = x

[H2] = 0.4815 + x

now

Kc = [C02] [H2] / [CO] [H20]

so

23.2 = [x] [0.4815 + x] / [0.03439 -x] [0.0535-x]

x = 0.0207137

so

[C02] = x = 0.0207137

now

moles = conc x volume

so

moles of C02 = 0.0207137 x 1.35 = 0.02796355

now

mass = moles x molar mass

so

mass of C02 = 0.02796355 x 44 = 1.23

so

1.23 grams of C02 is present at equilibrium