1.40g each of CO, H2O, and H2 are sealed in a 1.41−L vessel and brought to equilibrium at 600 K.

Question

CO(g)+H2O(g)⇌CO2(g)+H2(g)KC=23.2
How many grams of CO2 will be present in the equilibrium mixture?

Answer 1

(CO) = 1.40/molar mass CO = approx 0.05 mol and 0.05mol/1.41L = estd 0.03M

(H2O)= 1.4/molar mass H2O = approx 0.078 mols and 0.078mol/1.61L = 0.05M

(H2) = 1.4/molar mass H2 = approx 0.7 mols and 0.7 mol/1.61L = 0.4M

You need to do these more accurately as well as those that follow. I've estimated all of them.
..........CO(g)+H2O(g)⇌CO2(g)+H2(g)
I.......0.03...0.05.....0.....0.5
C.........-x....-x......x......+x
E......0.03-x..0.05-x...x.....0.5+x

Substitute the E line into the Kc expression and solve for x = (CO2) in M, then mols CO2 = M CO2 x L CO2
Finally, mols CO2 = grams/molar mass. You know mols and molar mass, solve for grams.