Chapter 15.3, Problem 2RQ

Interpretation Introduction

Interpretation: Volume of $Ba{\left(N{O}_3\right)}_2$ has to be determined in the precipitation reaction.

Concept introduction: 3 moles of $Ba{\left(N{O}_3\right)}_2$ reacts with 2 moles of $N{a}_{3}P{O}_4$ to produce 1 mole of $B{a}_3{(P{O}_4)}_2$ .

The balanced equation: $\text{3}Ba{\left(N{O}_3\right)}_2+2N{a}_{3}P{O}_4\to B{a}_3{(P{O}_4)}_2+6NaN{O}_3$

Answer to Problem 2RQ

Volume of $Ba{\left(N{O}_3\right)}_2$ is $167mL.$

Explanation of Solution

From the reaction equation it is obvious, 3 moles of $Ba{\left(N{O}_3\right)}_2$ = 2 moles of $N{a}_{3}P{O}_4$

  $\therefore$ Moles of $Ba{\left(N{O}_3\right)}_2$

  $=\left(\frac{2}{3}\right)\text{ moles of }N{a}_{3}P{O}_4$

Let “ $V_1$ ” is the volume of $Ba{\left(N{O}_3\right)}_2$ needed in the reaction.

  $\therefore$ $M_1{V}_1=\left(\frac{2}{3}\right)M_2{V}_2$ is the required equation.

• $M_1=\text{concentration of }Ba{\left(N{O}_3\right)}_2=0.10M$
• $M_2=\text{concentration of }N{a}_{3}P{O}_4=0.10M$
• $V_2=\text{volume of }N{a}_{3}P{O}_4=250.0mL$

Putting the appropriate values in the required equation we get,

  $\begin{array}{c}(0.10M)\times V_1=\left(\frac{2}{3}\right)\times (0.10M)\times \left(250.0mL\right)\\ V_1=\frac{16.7}{0.10}mL\\ V_1=167mL\end{array}$

Conclusion

Volume of $Ba{\left(N{O}_3\right)}_2$ is $167mL.$