How many grams of solid $BaS{O_4}$ will form when $N{a_2}S{O_4}$ reacts with $25ml$ of $0.50M$ $Ba{(N{O_3})_2}$ ? $\_Ba{(N{O_3})_2} + \_N{a_2}S{O_4} \to \_BaS{O_4} + \_NaN{O_3}$ .

Hint: In order to determine the weight of solid barium sulphate we have to calculate the number of moles of barium nitrate by the relation of concentration and number of moles. (As molarity is equal to the ratio of number of mole and volume of solution in litre)

Complete step by step answer:
The balanced chemical equation will be:
 \[Ba{(N{O_3})_2}_{(aq)} + N{a_2}S{O_4}_{(aq)} \to BaS{O_4}_{(s)} + 2NaN{O_{3(aq)}}\]
As we know that $Ba{(N{O_3})_2}$ , $N{a_2}S{O_4}$ and $NaN{O_3}$ can dissociate into their respective ions. Due to their dissociation, the net reaction will take place.
The dissociation takes place as:
$Ba_{(aq)}^{2 + } + SO_{4(aq)}^{2 - } \to BaS{O_{4(s)}}$
Sulphate compounds formed with barium ions are insoluble in water. In more precise words, they are very slightly soluble in water. The reaction given above in question is a double replacement reaction and as a product precipitate of barium sulphate formed.
From the balanced chemical equation, we can clearly say that one mole of barium nitrate is used to form one mole of barium sulphate.
So, the number of moles of barium sulphate formed will be calculated as:
Concentration = number of moles / volume of solution in litre
So, the number of moles is equal to the product of concentration and volume of solution.
$ \Rightarrow {n_{(Ba{{(N{O_3})}_2})}} = C \times v$
$ \Rightarrow {n_{(Ba{{(N{O_3})}_2})}} = 0.50M \times 25 \times {10^{ - 3}}l$
$ \Rightarrow {n_{(Ba{{(N{O_3})}_2})}} = 0.012$
As, the number of moles of barium nitrate is equal to the number of moles of barium sulphate formed.
So, we can write: ${n_{(BaS{O_4})}} = 0.012$
As molar mass of barium sulphate is $233.3$ .
That is one mole of barium sulphate weighs $233.3g$ so $0.012mole$ of barium sulphate will weigh:
$\
   \Rightarrow 0.012 \times 233.3g \\
   \Rightarrow 2.79g \\
    \\
\ $

Note:
Barium sulphate is a white crystalline powder. It is odorless and insoluble in water or we can say that it is slightly soluble in water. The solubility product of barium sulphate is very low.