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Stoichiometric Calculations

If 0.50 mol...

Question

If $0.50$ mol of $C a C l_{2}$ is mixed with $0.20$ mol of $N a_{3} P O_{4}$ , the maximum number of moles of $C a_{3} (P O_{4})_{2}$ 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 D $0.10$
The balanced chemical reaction is as shown below:
$3 B a C l_{2} + 2 N a_{3} P O_{4} \Rightarrow B a_{3} (P O_{4})_{2} + 6 N a C l$
$3$ moles of $B a C l_{2}$ combines with $2$ moles of $N a_{3} P O_{4}$ .
$0.5$ moles of $B a C l_{2}$ will combine with $\frac{2}{3} \times 0.5 = 0.33$ moles of $N a_{3} P O_{4}$ .
$N a_{3} P O_{4} (0.2$ mol) available is less than the required amount. It acts as a limiting reagent. It determines the amount of product.
$2$ moles of $N a_{3} P O_{4}$ react to form $1$ mole of $B a_{3} (P O_{4})_{2}$ .
$0.2$ moles of $N a_{3} P O_{4}$ will react to form $\frac{1}{2} \times 0.2 = 0.1$ mole of $B a_{3} (P O_{4})_{2}$ .

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