Question

When CO2(g) (0.118 mol/L) and H2(g) (3.90 mol) in a 33.0 L reaction vessel at 980K are allowed to come to equilibrium the mixture contains 0.0853 mol/L of H20(g). What concentration (mol/L) of CO2(g) reacted?

CO2(g) + H2(g) <--> CO(g) + H2O(g)

Answer 1

If you will do the ICE chart I think you can see the answer.
CO2 + H2 <==> H2O + CO

I= initial concn:
(CO2)=0.118 M
(H2) = 3.90 M if the problem means mols/L or 3.90/33 = 0.118 M if it means mols in 33.0 L.
(H2O) = 0
(CO) = 0

C = change in concn:
(H2O) = +x
(CO) = +x
(CO2) = - x
(H2) = -x

E = equilibrium concn:
(H2O) = 0.0853 from the problem.
(CO) we know from the equation that this is also 0.0853.
(CO2) = 0.118 - x but you know what x is. It is 0.0853.
Check my work. Check my thinking.