S2 + C ----> CS2How many grams of CS2 can be prepared by heating 11.0 moles of S2 with excess carbon in 5.20 L reaction vessel held at 900 kelvin until equilibrium is attaine…

Question

asked Sep 3, 2020 144k views

1 Answer

Jeremy Burton · Answer 1 · 2020-09-07T13:05:18+0000

Answer : The mass of
prepared can be, 754.832 grams

Explanation : Given,

Initial moles of
S_2 = 11.0 mole

Volume of solution = 5.2 L

Equilibrium constant
(K_c) = 9.40

First we have to calculate the concentration of
.

$\text{Concentration of }S_2=\frac{\text{Moles of }S_2}{\text{Volume of solution}}=(11.0mole)/(5.20L)=2.115M$

The balanced equilibrium reaction will be,

$S_2(g)+C(s)\rightleftharpoons CS_2(g)$

Initial moles 2.115 0

At eqm. (2.115-x) x

The equilibrium expression for this reaction will be,

K_c=([CS_2])/([S_2])

Now put all the given values in this expression, we get:

9.40=((x))/((2.115-x))

x=1.91M

The concentration of
CS_2 = x = 1.91 M

Now we have to calculate the moles moles of
.

Formula used :
Concentration=(Moles)/(Volume)

$\text{Concentration of }CS_2=\frac{\text{Moles of }CS_2}{\text{Volume of solution}}$

$1.91M=\frac{\text{Moles of }CS_2}{5.20L}$

$\text{Moles of }CS_2=9.932mole$

Now we have to calculate the mass of
.

$\text{Mass of }CS_2=\text{Moles of }CS_2* \text{Molar mass of }CS_2$

$\text{Mass of }CS_2=9.932mole* 76g/mole=754.832g$

Therefore, the mass of
prepared can be, 754.832 grams