Respuesta :
Answer : The mass of [tex]CS_2[/tex] prepared can be, 754.832 grams
Explanation : Given,
Initial moles of [tex]S_2[/tex] = 11.0 mole
Volume of solution = 5.2 L
Equilibrium constant [tex](K_c)[/tex] = 9.40
First we 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{11.0mole}{5.20L}=2.115M[/tex]
The balanced equilibrium reaction will be,
[tex]S_2(g)+C(s)\rightleftharpoons CS_2(g)[/tex]
Initial moles 2.115 0
At eqm. (2.115-x) x
The equilibrium expression 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)}{(2.115-x)}[/tex]
[tex]x=1.91M[/tex]
The concentration of [tex]CS_2[/tex] = x = 1.91 M
Now we have to calculate the moles moles of [tex]CS_2[/tex].
Formula used : [tex]Concentration=\frac{Moles}{Volume}[/tex]
[tex]\text{Concentration of }CS_2=\frac{\text{Moles of }CS_2}{\text{Volume of solution}}[/tex]
[tex]1.91M=\frac{\text{Moles of }CS_2}{5.20L}[/tex]
[tex]\text{Moles of }CS_2=9.932mole[/tex]
Now we have to calculate the mass of [tex]CS_2[/tex].
[tex]\text{Mass of }CS_2=\text{Moles of }CS_2\times \text{Molar mass of }CS_2[/tex]
[tex]\text{Mass of }CS_2=9.932mole\times 76g/mole=754.832g[/tex]
Therefore, the mass of [tex]CS_2[/tex] prepared can be, 754.832 grams
