Methanol, CH3OH(g), is manufactured from carbon momoixde, CO(g), and hydrogen, H2(g), according to the?

Question

following equation:

CO(g) + 2H2(g) = CH3OH(g)

A 1.00-L container is filled with 0.11 mol CO(g) and 0.200 mol H2(g), The reaction is allowed to proceed at 200'C. At equilibruim ther is 0.120 mol H2(g)
a) What are the equilibruim concentrations of CO(g) and CH3OH(g)?

b) Calculate the percent reaction

Answer 1

First of all this reaction is gas phase and not in aqueous medium as the guy above me has posted.

CO(g) + 2H2(g) = CH3OH(g)

At equilibrium 0.12 mol H2 is present.
H2 consumed = 0.2-0.12=0.08 mol
Per 2 mol of H2 consumed ,1mol of CO is also consumed and formation of 1 mol of CH3OH occurs.

For 0.08 mol H2 consumed, 0.04 mol CO is consumed and formation of 0.04 mol of CH3OH occurs.

Thus at EQULIBRIUM:
CO(g) = 0.11-0.04=0.07 mol

CH3OH(g) = 0.04 mol

100% reaction= complete conversion to CH3OH(g)
Since 0.200 mol H2(g) is the limiting reagent:
100% reaction = 0.1 mol CH3OH(g)

Since 0.04 mol is formed in equilibrium:
% reaction = (0.04/0.1)*100 = 40%

Answer 2

Assuming that they are put in quantities to make them 0.11 molar and 0.2 molar in 1 litre solution.

Then at equilibrium you have used 0.08 mols of H2

You need twice as much H2 as you do CO - so you have used 0.04 mols of CO

Therefore the concentration of CO is 0.11 - 0.04 = 0.07 mol CO(g)

The ratio of CO used is 1:1 with the amount of CH3OH (g) created so you have 0.04 mol CH30H (g)

The percentage reaction is based on the CO (g) (I'm guessing) so this will be 0.04/0.11 x 100 = 36.4%

If you are basing this on the H2 (g) then it will be 0.08/0.2 x 100 which = 40%

Hope that helps.