A mixture of 0.8 mol of CO(g) and 0.8 mol of H2(g) was placed in a sealed 1.0 L container - HSC - SSCE Chemistry - Question 7 - 2023 - Paper 1

To determine the amount of H2(g) present at equilibrium, we can use the stoichiometry of the reaction.

1. Initial Moles:

   • CO(g): 0.8 mol
   • H2(g): 0.8 mol
   • CH3OH(g): 0 mol

2. Change in Moles: At equilibrium, the amount of CO(g) is 0.5 mol. Therefore, the change in the amount of CO(g) is:

   $0.8 ext{ mol} - 0.5 ext{ mol} = 0.3 ext{ mol}$

   Since the reaction consumes 1 mol of CO(g) for every 2 mol of H2(g), then the change in moles of H2(g) consumed will be:

   $2 imes 0.3 ext{ mol} = 0.6 ext{ mol}$

3. Equilibrium Moles of H2(g):

   $0.8 ext{ mol} - 0.6 ext{ mol} = 0.2 ext{ mol}$

Thus, the amount of H2(g) present at equilibrium is 0.2 mol. Therefore, the answer is A. 0.2 mol.