Chemistry - Ionic Equilibrium Question with Solution | TestHub

$$3H_{2}S{O}_4+B{a}_3(P{O}_4{)}_2\to 3BaS{O}_4\downarrow +2H_{3}P{O}_4$$

$0.2\times 0.1$ and $0.4\times 0.05$

$0.02$ MOLE $0.02$ mole

Here, $H_{2}S{O}_4$ is the limiting reactant (LR).

Therefore, the remaining moles of $B{a}_3(P{O}_4{)}_2=0.02-\frac{0.02}{3}$

$=\frac{0.04}{3}$ MOLE

Thus, the number of moles of $B{a}^{2+}$ ions $=3\times \frac{0.04}{3}=0.04$ mole.

The volume of the solution $=0.1+0.4=0.5$ L.

Therefore, $M=\frac{0.04}{0.5}=0.08M=X$.

So, $100X=0.08\times 100=8$.