If 0.5 moles of $B a C l_{2}$ is mixed with 0.2 moles of $N a_{3} P O_{4}$ the maximum moles of $B a_{3} (P O_{4})_{2}$ obtained is

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Solution

$3 B a C l_{2} + 2 N a_{3} P O_{4} \to B a_{3} (P O_{4})_{2} + 6 N a C l$

Finding the limiting reagent:
$\frac{given moles of B a C l_{2}}{stoichiometric coefficient} = \frac{0.5}{3}$

$\frac{given moles of N a_{3} P O_{4}}{stoichiometric coefficient} = \frac{0.2}{2}$

Hence the limiting reagent is $N a_{3} P O_{4}$

From stoichiometry,
2 moles of $N a_{3} P O_{4}$ gives 1 mole of $B a_{3} (P O_{4})_{2}$
So, 0.2 mole give 0.1 mole of $B a_{3} (P O_{4})_{2}$

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