Does the number of moles of reaction products increase,decrease,or remain the same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?
$(a)$ $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$
$(b)$ $CaO_{(s)} + CO_{2(g)} \rightleftharpoons CaCO_{3(s)}$
$(c)$ $3Fe_{(s)} + 4H_{2}O_{(g)} \rightleftharpoons Fe_{3}O_{4(s)} + 4H_{2(g)}$

(N/A) According to Le Chatelier's principle,when the pressure of a system at equilibrium is decreased (by increasing the volume),the equilibrium shifts in the direction that increases the total number of moles of gaseous species.
$(a)$ $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$: The number of gaseous moles increases from $1$ (reactant) to $2$ (products). Thus,the equilibrium shifts forward,and the number of moles of products increases.
$(b)$ $CaO_{(s)} + CO_{2(g)} \rightleftharpoons CaCO_{3(s)}$: The number of gaseous moles is $1$ on the reactant side and $0$ on the product side. To increase gaseous moles,the equilibrium shifts in the reverse direction,so the number of moles of products decreases.
$(c)$ $3Fe_{(s)} + 4H_{2}O_{(g)} \rightleftharpoons Fe_{3}O_{4(s)} + 4H_{2(g)}$: The number of gaseous moles is $4$ on both sides. Therefore,a change in pressure has no effect on the equilibrium position,and the number of moles of products remains the same.

Reaction	Change in Moles of Gas (Products vs Reactants)
$(a)$	Products: $2$,Reactants: $1$. Equilibrium shifts forward.
$(b)$	Products: $0$,Reactants: $1$. Equilibrium shifts reverse.
$(c)$	Products: $4$,Reactants: $4$. No change in equilibrium.