If $0.5$ mole of $B a C l_{2}$ is mixed with $0.2$ mole of $N a_{3} P O_{4}$ , the maximum number of moles of $B a_{3} (P O_{4})_{2}$ that can be formed is

0.70

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0.50

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0.20

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0.10

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Solution

The correct option is D $0.10$
The balanced reaction is
$3 B a C l_{2} + 2 N a_{3} P O_{4} \to B a_{3} (P O_{4})_{2} + 6 N a C l$
In this reaction, $3$ moles of $B a C l_{2}$ combines with $2$ moles of $N a_{3} P O_{4}$ .
Hence $0.5$ mole of $B a C l_{2}$ recquire $= \frac{2}{3} \times 0.5 = 0.33$ moles of $N a_{3} P O_{4}$ .
Since, available $N a_{3} P O_{4}$ $(0.2$ mole $)$ is less than required mole $(0.33)$ , it is the limiting reactant and would determine the amount of product $B a_{3} (P O_{4})_{2}$ .
$∴ 2$ moles of $N a_{3} P O_{4}$ gives $1$ mole of $B a_{3} (P O_{4})_{2}$ .
$∴ 0.2$ mole of $N a_{3} P O_{4}$ would give $= \frac{1}{2} \times 0.2 = 0.1$ mole $B a_{3} (P O_{4})_{2}$

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