Question

Methanol, CH3OH, formerly known as wood alcohol, is manufactured commercially by the following reaction A 1.500-L vessel was filled with 0.1500 mol CO and 0.3000 mol H2. When this mixture came to equilibrium at 500 K, the vessel contained 0.1187 mol CO. How many moles of each substance were in the vessel?

Answer 1

Answer : The moles of $CO,H_2\text{ and }CH_3OH$ are, 0.1187, 0.2374 and 0.0313 moles respectively.

Explanation : Given,

Initial moles of $CO$ = 0.15 mole

Initial moles of $H_2$ = 0.3 mole

Moles of $CO$ at equilibrium = 0.1187 mole

The given equilibrium reaction is,

                             $CO+2H_2\rightleftharpoons CH_3OH$

Initial moles          0.15    0.3          0

At equilibrium  (0.15-x)  (0.3-2x)    x

As we are given that,

Moles of $CO$ at equilibrium = 0.1187 mole = 0.15 - x

0.15 - x = 0.1187

x = 0.0313

Moles of $H_2$ at equilibrium = 0.3 - 2x = 0.3 - 2(0.0313) = 0.2374 mole

Moles of $CH_3OH$ at equilibrium = x = 0.0313 mole

Therefore, the moles of $CO,H_2\text{ and }CH_3OH$ are, 0.1187, 0.2374 and 0.0313 moles respectively.