Question

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 attained?Kc= 9.40

Answer 1

Answer : The mass of
`CS_2` 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
`S_2`.

`\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
`CS_2`.

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
`CS_2`.

`\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
`CS_2` prepared can be, 754.832 grams