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 12.5 mol S2(g) with excess carbon in a 5.30 L reaction vessel held at 900 K until equilibrium is attained?

Answer 1

Answer: The amount of carbon sulfide prepared is 859.1724 grams.

Step-by-step explanation:

To calculate the concentration of
`S_2`, we use the equation:

`\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}` .....(1)

We are given:

Moles of
`S_2` = 12.5 moles

Volume of solution = 5.3 L

Putting values in above equation, we get:

`\text{Initial concentration of }S_2=(1.25mol)/(5.3L)=2.36M`

For the reaction of sulfur and carbon, the equation follows:

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

At
`t=0` 2.36

At
`t=t_(eq)` 2.36 - x x

• The expression for equilibrium constant for the above reaction follows:

`K_c=(CS_2)/(S_2)`

We are given:

`K_c=9.40`

`[CS_2]=x`

`[S_2]=2.36-x`

Putting values in above equation, we get:

`9.40=(x)/(2.36-x)\\\\x=2.133`

• Now, calculating the moles of carbon sulfide using equation 1, we get:

Molarity of
`CS_2` = 2.133 M

Volume of solution = 5.30 L

Putting values in equation 1, we get:

`2.133mol/L=\frac{\text{Moles of }CS_2}{5.30L}\\\\\text{Moles of }CS_2=11.3049mol`

• To calculate the mass of carbon sulfide, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

Moles of carbon sulfide = 11.3049 mol

Molar mass of carbon sulfide = 76 g/mol

Putting values in above equation, we get:

`11.3049mol=\frac{\text{Mass of carbon sulfide}}{76g/mol}\\\\\text{Mass of carbon sulfide}=859.1724g`

Hence, the amount of carbon sulfide prepared is 859.1724 grams.