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Ideal Gas Equation

0.3 moles of ...

Question

$0.3$ moles of $B a C l_{2}$ is mixed with $0.2$ 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:

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

0.7

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0.5

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0.3

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0.1

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Solution

The correct option is C $0.1$
$3 B a C l_{2} + 2 N a_{3} P O_{4} ⟶ B a_{3} (P O_{4})_{2} + 6 N a C l_{2}$
$0.3 m o l e s$ $0.2 m o l e s$

$3$ moles of $B a C l_{2}$ reacts with $2$ moles of $N a_{3} P O_{4}$ to give $1$ mole of $B a_{3} (P O_{4})_{2}$ .

$∴$ $0.3$ moles of $B a C l_{2}$ react with $0.2$ moles of $N a_{3} P O_{4}$ to give $0.1$ mole of $B a_{3} (P O_{4})_{2}$ .

Hence, option $D$ is correct.

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