What is the maximum mass of ${Ba}_{3} {({PO}_{4})}_{2}$ that can be formed from 0.00240 mol of $Ba {({NO}_{3})}_{2}$ and $0.131 g$ of ${Na}_{3} {PO}_{4}$ ?

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1.44 g

7.22 g

0.480 g

0.240 g

answer is A.

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Detailed Solution

In the reaction we have

$n = \begin{array}{ccc} 3 Ba {({NO}_{3})}_{2} + 2 {Na}_{3} {PO}_{4} \to & {Ba}_{3} {({PO}_{4})}_{2} \\ 0.0024 & 0.131 / 164 \\ (LR) 4 \times 10^{- 4} \times 601 = 0.240 g \end{array}$

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