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?

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Alleei Alleei · Answer 1 · 2019-08-01T08:19:43+02:00

Answer : The mass of [tex]CS_2[/tex] is, 555.028 grams

Explanation :

First er have to calculate the concentration of [tex]S_2[/tex].

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

Now we have to calculate the concentration of [tex]CS_2[/tex].

The given balanced chemical reaction is,

[tex]S_2(g)+C(s)\rightleftharpoons CS_2(g)[/tex]

Initial conc. 1.51 0 0

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

The expression for equilibrium constant for this reaction will be,

[tex]K_c=\frac{[CS_2]}{[S_2]}[/tex]

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

[tex]9.40=\frac{x}{(1.51-x)}[/tex]

By solving the term 'x', we get :

x = 1.365 M

Concentration of [tex]CS_2[/tex] = x M = 1.365 M

Now we have to calculate the moles of [tex]CS_2[/tex].

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

Now we have to calculate the mass of [tex]CS_2[/tex].

Molar mass of [tex]CS_2[/tex] = 76 g/mole

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

Therefore, the mass of [tex]CS_2[/tex] is, 555.028 grams