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Oxidation Number Method

If 0.5 mol ...

Question

If $0.5$ mol of $B a C l_{2}$ is mixed with $0.2$ mol 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 A $0.10$
$3 B a C l_{2} + 2 N a_{3} P O_{4} \to B a_{3} (P O_{4})_{2} + 6 N a C l$

The coefficients of the balanced equation tell you the ratio of the reactants and the products.
3 moles of $B a C l_{2}$ react with 2 moles of $N a_{3} P O_{4}$ to yield 1 mole of $B a_{3} (P O_{4})_{2}$ .

Required ratio of $B a C l_{2} : N a_{3} P O_{4} = 3 : 2$ .

0.5 mole $B a C l_{2}$ requires 2/3 x 0.5 mole (0.333 mole) $N a_{3} P O_{4}$ .
You have only 0.2 mole $N a_{3} P O_{4}$ so it is limiting and $B a C l_{2}$ is in excess.

2 moles of $N a_{3} P O_{4}$ will yield 1 mole of $B a_{3} (P O_{4})_{2}$ .
0.2 mole $N a_{3} P O_{4}$ will yield 0.1 mol of $B a_{3} (P O_{4})_{2}$ .

D: 0.1 mol yield.

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