Question

Pcl5 => pcl3+cl2

Answer 1

10dm^3. At 250°c kc= 0.041

Answer 2

Answer: 0.47 moles

Explanation:

Initial moles of  $PCl_5$ = 1 mole

Volume of container = $10dm^3=10L$

Initial concentration of $PCL_5=\frac{moles}{volume}=\frac{1mole}{10L}=0.1M$  

The given balanced equilibrium reaction is,

                $PCL_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$

Initial conc.         0.1 M             0    0

At eqm. conc.    (0.1-x) M   (x) M   (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}$

$0.041=\frac{x\times x}{0.1-x}$

Solving for x:

$x=0.047$

Thus concentration of $Cl_2$ produced is 0.047 M

$Molarity=\frac{n}{V_s}$

where,

n= moles of solute  

$V_s$ = volume of solution in L

$0.047M=\frac{n}{10L}$

$n=0.47moles$

Thus moles of $Cl_2$ produced is 0.47.