Question

Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equation is S2(g)+C(s)↽−−⇀CS2(g)????c=9.40 at 900 K How many grams of CS2(g) can be prepared by heating 8.08 mol S2(g) with excess carbon in a 5.35 L reaction vessel held at 900 K until equilibrium is attained?

Answer 1

Answer : The mass of $CS_2$ is, 555.028 grams

Explanation :

First er have to calculate the concentration of $S_2$.

$\text{Concentration of }S_2=\frac{\text{Moles of }S_2}{\text{Volume of solution}}=\frac{8.08mole}{5.35L}=1.51mole/L$

Now we have to calculate the concentration of $CS_2$.

The given balanced chemical reaction is,

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

Initial conc.         1.51       0         0

At eqm. conc.   (1.51-x)  (x)       (x)

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[CS_2]}{[S_2]}$

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

$9.40=\frac{x}{(1.51-x)}$

By solving the term 'x', we get :

x = 1.365 M

Concentration of $CS_2$ = x M = 1.365 M

Now we have to calculate the moles of $CS_2$.

$\text{Moles of }CS_2=\text{Concentration of }CS_2}\times \text{Volume of solution}=1.365mole/L\times 5.35L=7.303mole$

Now we have to calculate the mass of $CS_2$.

Molar mass of $CS_2$ = 76 g/mole

$\text{Mass of }CS_2=\text{Moles of }CS_2}\times \text{molar mass of }CS_2}=7.303mole\times 76g/mole=555.028g$

Therefore, the mass of $CS_2$ is, 555.028 grams