Q. If 0.50 mole of $BaCl_{2}$ is mixed with 0.20 mol of $Na_{3}PO_{4}$ , the maximum number of moles of $Ba_{3}\left(\right.PO_{4}\left(\left.\right)_{2}$ that can be formed is

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A

0.10

B

0.20

C

0.30

D

0.40

Solution:

$\underset{0.5}{3 \left(\text{BaCl}\right)_{2}} + \underset{0.2}{2 \left(\text{Na}\right)_{3} \left(\text{PO}\right)_{4}} \rightarrow \left(\text{Ba}\right)_{3} \left(\right. \left(\text{PO}\right)_{4} \left.\right)_{2} + 6 \text{NaCl}$
Here Limiting reagent is $Na_{3}PO_{4}$
As 2 mole of $Na_{3}PO_{4}$ gives = 1 mol of $Ba_{3}\left(\right.PO_{4}\left(\left.\right)_{2}$ .
So (0.2 mol) of $Na_{3}PO_{4}$ gives = 0.1 mol of $Ba_{3}\left(\right.PO_{4}\left(\left.\right)_{2}$ .

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