carbon disulfide is prepared by heating sulfur and charcoal. the chemical equation is S2(g)+C(S)=CS2(g) and Kc = 9.4 at 900K.

How many grams of CS2(g) can be prepared by heating 8.9moles of S2(g) with excess carbon in a 5.00L reaction vessel held at 900K until equilibrium is attained?

1 answer

(S2) = mols/L = 8.9/5.00 = approx 1.8 but you need a better number than that. Recalculate all that follows.
......S2 + C(s) ==> CS2
I....1.8............0
C.....-x............x
E....1.8-x..........x
Note that C does not appear in the Kc expression because it is a solid.
Kc = (CS2)/(S2) = 9.4
Substitute and solve for x. I get approx 2, then mols = M x L = 5*2 = about 10 and grams = mols x molar mass CS2. Remember all the number I have placed above are estimates. Post your work if you get stuck.