If 5 moles of $C a C l_{2}$ is mixed with 2 moles of $N a_{3} P O_{4}$ the maximum number of moles of $C a_{3} (P O_{4})_{2}$ formed will be:
$C a C l_{2} + N a_{3} P O_{4} \to C a_{3} (P O_{4})_{2} + N a C l$

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Solution

Balanced chemical equation:
$3 C a C l_{2} + 2 N a_{3} P O_{4} \to C a_{3} (P O_{4})_{2} + 6 N a C l$
Finding limiting reagent:
$\frac{moles}{stoichiometric coefficient} = \frac{moles of calcium chloride}{3}$
$= \frac{5}{3} = 1.66$
$\frac{moles of sodium phosphate}{2} = \frac{2}{2} = 1$
Hence sodium phosphate is the limiting reagent.
$\frac{moles of sodium phosphate}{2} = \frac{moles of calcium phosphate}{1}$
Moles of calcium phosphate formed = 1 mol

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If 5 moles of CaCl2 is mixed with 2 moles of Na3PO4 the maximum number of moles of Ca3(PO4)2 formed will be: CaCl2+Na3PO4→Ca3(PO4)2+NaCl

If 5 moles of $C a C l_{2}$ is mixed with 2 moles of $N a_{3} P O_{4}$ the maximum number of moles of $C a_{3} (P O_{4})_{2}$ formed will be:
$C a C l_{2} + N a_{3} P O_{4} \to C a_{3} (P O_{4})_{2} + N a C l$