1. The problem statement, all variables and given/known data CH4 (g) + H2O (g) <- -> CO (g) + 3 H2 (g) .2 M CH4 and .2 M H2O are added to an empty 4 L sealed vessel at 1400 K. When equilibrium is established, the concentration of H2 is .444 M. Calculate Kc... which I did and got 4.8 which is the given answer. The real problem I have is part b) What mass of H2O (g) must be added to the equilibrium mixture in order to raise the concentration of H2 to .54 mol/L? The answer is supposed to be "20. g" 3. The attempt at a solution using the equilibrium concentrations I calculated for the first part as initial concentrations CH4 H2O CO 3H2 I .052 .052 .148 .444 C -.032 -.032 +.032 +.096 E .02 .02 .18 .54 So to change the .444 M H2 to .53 M... the reactants would need a third more added to them? and .032 M/L in 4.0 L would be 0.128 mol... if the molar mass of H2O is 18.0152 then that would be 2.3 g... which is nowhere near 20 g Am I way off? I'm confused.