Question

$0.2mol$ of $N{a}_{3}P{O}_4$ and $0.5mol$ of $Ba(N{O}_3{)}_2$ are mixed in a $1L$ of solution. Which of the following statements is/are correct?

• A. $0.2mol$ of barium phosphate precipitate is obtained
• B. $0.1mol$ of barium phosphate precipitate is obtained
• C. Molarity of $B{a}^{2+}$ ions in the resulting solution is $0.2$
• D. Molarities of $N{a}^{+}$ and $N{O}_3^{-}$ ions are $0.6$ and $1.0$ respectively

Answer 1

Answer: (B), (C), (D)

The correct options are
B $0.1mol$ of barium phosphate precipitate is obtained
C Molarity of $B{a}^{2+}$ ions in the resulting solution is $0.2$
D Molarities of $N{a}^{+}$ and $N{O}_3^{-}$ ions are $0.6$ and $1.0$ respectively

The corresponding balanced reaction is:
$2N{a}_{3}P{O}_4\left(aq\right)+3Ba\left(N{O}_3{)}_2\right(aq)\to B{a}_3(P{O}_4{)}_2\left(aq\right)\downarrow +6NaN{O}_3\left(aq\right)$
So, $N{a}_{3}P{O}_4$ is the limiting reagent and is completely consumed.
Moles of $B{a}_3(P{O}_4{)}_2$ formed $=\frac{0.2}{2}mol=0.1mol$
Moles of unreacted $Ba(N{O}_3{)}_2=(0.5-0.3)mol=0.2=\text{mol of}B{a}^{2+}ion$.
The reacted 0.3 $B{a}^{2+}$ will be precipitated as 0.1 mole barium phosphate.
Moles of $N{a}^{+}$ in solution $=0.2\times 3=0.6$
Moles of $N{O}_3^{-}$ in solution $=0.5\times 2=1$