A) 4.2 moles
B) 2.1 moles
C) 5.5 moles
D) 6.3 moles
Correct Answer: B
Solution :
[b] At equilibrium the moles of \[C{{l}_{2}}\] must be \[=0.15\times 3=0.45\] \[\begin{align} & \begin{matrix} {} & {} & {} & \,\,\,\,\,\,\,\,\,PC{{l}_{5}}\rightleftharpoons PC{{l}_{3}}+C{{l}_{2}} \\ \end{matrix} \\ & \begin{matrix} \text{Eqm}\text{.}\,\,\text{Conc}\text{.} & \frac{x-0.45}{3} & \frac{0.45}{3} & \frac{0.45}{3} \\ \end{matrix} \\ \end{align}\] \[{{K}_{c}}=\frac{\left[ PC{{l}_{3}} \right]\left[ C{{l}_{2}} \right]}{\left[ PC{{l}_{5}} \right]}\] \[\therefore 0.04=\frac{0.15\times 0.15}{(x-0.45)/3}\] \[\therefore \text{ }x=2.1\text{ }moles\]
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