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Ideal Gas Equation

If 0.5 mole o...

Question

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

A

0.7 mol

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B

0.5 mol

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C

0.2 mol

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D

0.1 mol

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Solution

The correct option is D 0.1 mol
$B a C l_{2} + N a_{3} P O_{4} \to B a_{3} (P O_{4})_{2} + N a C l$
Equivalents of $B a C l_{2} = 0.5 \times n_{f} = 0.5 \times 2 = 1$
Equivalents of $N a_{3} P O_{4} = 0.2 \times n_{f} = 0.5 \times 3 = 0.6$
Equivalents of $B a_{3} (P O_{4})_{2} f o r m e d = 0.6 = m o l e s \times n_{f}$
Moles of $B a_{3} (P O_{4})_{2} = \frac{0.6}{6} = 0.1 m o l e s$

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