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Potassium Permanganate

If 0.5 mole...

Question

If $0.5$ moles of $B a C l_{2}$ is mixed with $0.20$ moles 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 :

A

0.1

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B

0.2

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C

0.5

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D

0.7

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Solution

The correct option is C $0.1$
$3 B a C l_{2} 3 m o l + 2 N a_{3} P O_{4} 2 m o l \to 6 N a C l 6 m o l + B a_{3} (P O_{4})_{2} 1 m o l$
Given, $0.5$ mol of $B a C l_{2}$ and $0.2$ mol of $N a_{3} P O_{4}$ are mixed.
$2$ mol of $N a_{3} P O_{4} \Rightarrow$ $3$ mol of $B a C l_{2}$
$0.2$ mol of $N a_{3} P O_{4} \Rightarrow$ $0.3$ mol of $B a C l_{2}$
$∴ N a_{3} P O_{4}$ is the limiting reagent.
$∴$ $2$ mol of $N a_{2} P O_{4} \Rightarrow$ $1$ mol of $B a_{3} (P O_{4})_{2}$
$0.2$ mol of $N a_{3} P O_{4} \Rightarrow$ $0.1$ mol of $B a_{3} (P O_{4})_{2}$

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