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Redox Reactions

If 0.5 moles ...

Question

If 0.5 moles of $B a C l_{2}$ is mixed with 0.1 moles of $N a_{3} P O_{4},$ the maximum amount of $B a_{3} (P O_{4})_{2}$ that can be formed is

0.7 m o l

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0.5 m o l

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0.2 m o l

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0.05 m o l

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Solution

The correct option is D $0.05 m o l$
Let us first solve this problem by writing the complete balanced reaction.
$3 B a C l_{2} + 2 N a_{3} P O_{4} \to B a_{3} (P O_{4})_{2} ↓ + 6 N a C l$

We can see that the moles of $B a C l_{2}$ used are $\frac{3}{2}$ times the moles of $N a_{3} P O_{4} .$

Therefore, to react with $0.1 m o l$ of $N a_{3} P O_{4},$ the moles of $B a C l_{2}$ required would be $0.1 \times \frac{3}{2} = 0.15.$

Since the amount of $B a C l_{2}$ is $0.5 m o l$ , we can conclude that $N a_{3} P O_{4},$ is the limiting reagent.

Therefore, moles of $B a_{3} (P O_{4})_{2}$ formed are $0.1 \times \frac{1}{2} = 0.05 m o l .$
Hence, (d) is the correct answer.

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