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?
$(a) \quad PCl_{5(g)} \longleftrightarrow PCl_{3(g)} + Cl_{2(g)}$
$(b) \quad CaO_{(s)} + CO_{2(g)} \longleftrightarrow CaCO_{3(s)}$
$(c) \quad 3Fe_{(s)} + 4H_2O_{(g)} \longleftrightarrow Fe_3O_{4(s)} + 4H_{2(g)}$

(N/A) According to Le Chatelier's principle,decreasing the pressure (by increasing the volume) shifts the equilibrium toward the side with a greater number of moles of gaseous species.
$(a)$ $PCl_{5(g)} \longleftrightarrow PCl_{3(g)} + Cl_{2(g)}$: There are $1$ mole of gas on the reactant side and $2$ moles of gas on the product side. Since the product side has more moles,the equilibrium shifts forward,and the number of moles of products increases.
$(b)$ $CaO_{(s)} + CO_{2(g)} \longleftrightarrow CaCO_{3(s)}$: There is $1$ mole of gas on the reactant side and $0$ moles of gas on the product side. Since the reactant side has more moles,the equilibrium shifts backward,and the number of moles of products decreases.
$(c)$ $3Fe_{(s)} + 4H_2O_{(g)} \longleftrightarrow Fe_3O_{4(s)} + 4H_{2(g)}$: There are $4$ moles of gas on the reactant side and $4$ moles of gas on the product side. Since the number of moles of gas is equal on both sides,the equilibrium position remains unchanged,and the number of moles of products remains the same.