Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equati
ID: 1065829 • Letter: C
Question
Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equation is S_2 (g) + C(s) equilibrium CS_2 (g) Kc = 9.40 at 900 K How many grams of CS_2 (g) can be prepared by heating 14.2 moles of S_2 (g) with excess carbon in a 7.50 L reaction vessel held at 900 K until equilibrium is attained? Start by finding the initial concentration of S_2 using the number of moles and the volume. Then, set up a table of initial and final concentrations, where the change in concentration of the product is designated as x.Explanation / Answer
initial concentration of S2 = moles/volume
=14.2 mol / 7.50 L
=1.893 M
S2 (g) + C (s) <----------> CS2 (g)
1.893 0 (initial)
1.893-x x (at equilibrium)
Kc = [CS2] / [S2]
9.40 = x / (1.893-x)
17.8 - 9.40*x = x
x = 1.71 M
[CS2] = 1.71 M
moles of CS2 = concentration * volume
=1.71 * 7.50
= 12.83 mol
mass of CS2 = number of mol * molar mass
= 12.83 mol * 76.14 g/mol
=976.9 g
Answer: 976.9 g