Respuesta :
Answer: 0.47 moles
Explanation:
Initial moles of [tex]PCl_5[/tex] = 1 mole
Volume of container = [tex]10dm^3=10L[/tex]
Initial concentration of [tex]PCL_5=\frac{moles}{volume}=\frac{1mole}{10L}=0.1M[/tex]
The given balanced equilibrium reaction is,
[tex]PCL_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)[/tex]
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,
[tex]K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}[/tex]
[tex]0.041=\frac{x\times x}{0.1-x}[/tex]
Solving for x:
[tex]x=0.047[/tex]
Thus concentration of [tex]Cl_2[/tex] produced is 0.047 M
[tex]Molarity=\frac{n}{V_s}[/tex]
where,
n= moles of solute
[tex]V_s[/tex] = volume of solution in L
[tex]0.047M=\frac{n}{10L}[/tex]
[tex]n=0.47moles[/tex]
Thus moles of [tex]Cl_2[/tex] produced is 0.47.
