Question

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

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

How many grams of CS2(g) can be prepared by heating 12.2 moles of S2(g) with excess carbon in a 7.40 L reaction vessel held at 900 K until equilibrium is attained?

Answer 1

Initial concentration of S2 = mol of S2 / volume in L

= 12.2 mol / 7.40 L

= 1.65 M

ICE Table:

Equilibrium constant expression is

Kc = [CS2]/[S2]

9.4 = (1*x)/((1.65-1*x))

15.51-9.4*x = 1*x

15.51-10.4*x = 0

x = 1.49

At equilibrium:

[CS2] = x = 1.49 M

volume , V = 7.4 L

use:

number of mol,

n = Molarity * Volume

= 1.49*7.4

= 11.03 mol

Molar mass of CS2,

MM = 1*MM(C) + 2*MM(S)

= 1*12.01 + 2*32.07

= 76.15 g/mol

use:

mass of CS2,

m = number of mol * molar mass

= 11.03 mol * 76.15 g/mol

= 840 g

Answer: 840 g