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.

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:

$\text{Initial concentration of }S_2=\frac{1.25mol}{5.3L}=2.36M$

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:

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

We are given:

$K_c=9.40$

$[CS_2]=x$

$[S_2]=2.36-x$

Putting values in above equation, we get:

$9.40=\frac{x}{2.36-x}\\\\x=2.133$

• 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.