Question

Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equation is

How many grams of CS2(g) can be prepared by heating 13.0 moles of S2(g) with excess carbon in a 8.15 L reaction vessel held at 900 K until equilibrium is attained?

S2(g)+C(s)<--------->CS2(g) Kc=9.40 at 900 K

__________g

Answer 1

The given reaction is

C(s) + S2 (g) <----------> CS2 (g)

we can write equilibrium constant Kc as

Kc = [CS2]/ [S2]

Let us draw an ICE table for the above reaction

C(s) + S2 (g) <----------> CS2 (g)

	C(s)	S2(g)	CS2 (g)
I	-	13.0 mol	0
C	-	-x	+x
E	-	13.0 - x	x

Substituting E values in equilibrium constant expression, we get

Kc = x / 13 - x

But Kc = 9.4

9.4 = x/13- x

9.4 ( 13 - x) = x

122.2 - 9.4 x - x = 0

10.4 x = 122.2

x = 11.75 mol

Moles of CS2 formed at equilibrium are 11.75 mol

Gram of CS2 can be calculated as 11.75 mol CS2 * 76.14 g/ 1 mol = 894. 6 g CS2 ~ 895 g CS2

895 g of CS2 can be prepared until equilibrium is attained