Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equation is S2 +C =CS2

Kc= 9.40 at 900k

How many grams of CS2(g) can be prepared by heating 14.3 moles of S2(g) with excess carbon in a 8.55 L reaction vessel held at 900 K until equilibrium is attained?

2 answers

............S2(g) + C(s) ==> CS2(g)
I..........1.67.....--.......0
C...........-x.......--......x
E.........1.67-x....--.......x

Note: (S2) = 14.3mols/8.55L = 1.6725 M.

Kc = 9.40 = (CS2)/(S2)
Substitute the equilibrium line into Kc expression and solve for x = (CS2) in mols/L.
x mols/L x 8.55 L = mols CS2
g CS2 = mols x molar mass.
.022