Question

How many grams of CS2(g) can be prepared by heating 16.4 moles of S2(g) with excess carbon in a 6.65 L reaction vessel held at 900 K until equilibrium is attained? Kc= 9.40 at 900k

Answer 1

The reaction is S2(g)+C ------->CS2(g)

moles of CS2= 16.4, concentration of CS2= moles/ unit volume= 16.4/6.65=2.5

let x= drop in concentration of S2 to reach equilibrium, At Equilibrium [CS2]= x and [S2] = 2.5-x

Kc= [CS2]/[S2] = x/(2.5-x)= 9.4

x= 2.5*9.4-9.4x

10.4x= 2.5*9.4

x= 2.3, hence at equilibrium, concentraton of CS2(g)= 2.3M, moles of CS2= molarity* Volume in L= 2.3*6.65=14.95 molar mass of CS2= 12+2*32= 76 g/mole, mass of CS2 formed = moles* molar mass= 14.95* 76 gm = 1136 gm