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Atomic Mass

If 0.5 mol of...

Question

If 0.5 mol of $B a C l_{2}$ is mixed with 0.20 mol of $N a_{3} P O_{4}$ , the maximum amount of $B a_{3} (P O_{4})_{2}$ that can be formed is :

0.70 mol

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0.50 mol

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0.20 mol

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0.10 mol

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Solution

The correct option is D 0.10 mol
$3 B a C l_{2} + 2 N a_{3} P O_{4} \to B a_{3} (P O_{4})_{2} + 6 N a C l$

Equivalents of $B a C l_{2} = m o l e s \times n_{f} = 0.5 \times 2 = 1$

Equivalents of $N a_{3} P O_{4} = m o l e s \times n_{f} = 0.2 \times 3 = 0.6$

Equivalents of $B a_{3} (P O_{4})_{2}$ formed $= 0.6 = m o l e s \times n_{f}$

Moles of $B a_{3} (P O_{4})_{2}$ formed $= \frac{0.6}{n_{f}} = \frac{0.6}{6} = 0.1 m o l$

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