Carbon monoxide and hydrogen react according to the equation:

CO(g) + 3H2(g) CH4(g) + H2O
When 1.00 mol CO and 3.00 mol H2 are placed in a 1L flask at 1200K and allowed to come to equilibrium, the reaction mixture is found to contain 0.387 mol H2O. What is the molar composition of the equilibrium mixture?

Well, if you have .387mol steam, then you used .387 mol CO, and 3*.387 mole H2. To find out the molar concentration of CO and H2, subtract what was used from the initial. The amount of CH4 of course is the same as H2O.

Moles of h2o driven off

And convert grams 0.85 to moles

To find the molar composition of the equilibrium mixture, we need to calculate the number of moles of each substance present. Let's break down the problem and solve it step by step:

Given:
1. Reaction equation: CO(g) + 3H2(g) → CH4(g) + H2O
2. Initial moles of CO = 1.00 mol
3. Initial moles of H2 = 3.00 mol
4. Moles of H2O in the equilibrium mixture = 0.387 mol

Step 1: Determine the change in moles:
According to the balanced equation, for every mole of CO that reacts, one mole of H2O is produced. Therefore, the number of moles of CO and H2O will be the same at equilibrium.

Change in moles of CO = -0.387 mol
Change in moles of H2 = +0.387 mol

Step 2: Calculate the moles of the remaining substances at equilibrium:
Moles of CO at equilibrium = Initial moles of CO + Change in moles of CO
Moles of CO at equilibrium = 1.00 mol - 0.387 mol = 0.613 mol

Moles of H2 at equilibrium = Initial moles of H2 + Change in moles of H2
Moles of H2 at equilibrium = 3.00 mol + 0.387 mol = 3.387 mol

Step 3: Use the balanced equation to find moles of CH4 at equilibrium:
Since one mole of CO reacts to produce one mole of CH4, the moles of CH4 will be the same as the moles of CO at equilibrium.

Moles of CH4 at equilibrium = Moles of CO at equilibrium = 0.613 mol

Step 4: Calculate the remaining moles of H2O at equilibrium:
Since one mole of CO reacts to produce one mole of H2O, and the change in moles of H2O is 0.387 mol, the moles of H2O at equilibrium will be the sum of the initial moles of H2O (0.387 mol) and the change in moles of H2O.

Moles of H2O at equilibrium = Initial moles of H2O + Change in moles of H2O
Moles of H2O at equilibrium = 0.387 mol + 0.387 mol = 0.774 mol

Now we have the moles of each substance at equilibrium:
CO: 0.613 mol
H2: 3.387 mol
CH4: 0.613 mol
H2O: 0.774 mol

Therefore, the molar composition of the equilibrium mixture is:
CO: 0.613 mol
H2: 3.387 mol
CH4: 0.613 mol
H2O: 0.774 mol