How many grams of CS2 can be prepared by heating 11.4 miles of S2 with excess ca

Question

How many grams of CS2 can be prepared by heating 11.4 miles of S2 with excess carbon in a 6.40L reaction vessel held at 900K until equilibrium is reached? Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equation is K:-9.40 at 900 K How many grams of CS(g) can be prepared by heating 11.4 moles of Sa(g) with excess carbon in a 6.40 L reaction vessel held at 900 K until equilibrium is attained? Number Exit". Check Answer 0 Next O Previous Give Up & View Solution contact us help terms

Explanation / Answer

For reaction:

S2 + C <------------> CS2

Initally S2 = 11.4 mol

Volume = 6.40 L

Molarity of CS2 = Number of moles/volume = 11.4/6.40 = 1.78 M

As C is in excess, so same number of moles of C will react.

Similarly, molarity of C = 1.78 M

ICE table for the reaction:

[C] [S2] [CS2]

Initial 1.78 M 1.78 M 0

Change -x -x +x

Equilibrium 1.78-x 1.78-x x

Equilibrium constant = Kc = [CS2]/([C][S2])

9.40 = x/(1.78-x)(1.78-x)

9.40 = x/(1.78-x)2

(1.78-x)2 = x/9.40

3.17+x2-3.56x = 0.106x

x2-3.67x+3.17 = 0

After solving quadratic equation, x = 1.4

So, concentration of [CS2] = x = 1.4 M

Moles of CS2 = molarity*volume = 1.4*6.4 = 8.96 moles

Moles of CS2 at equilibrium = 8.96 moles

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.