Question

How many grams of barium sulfate solid are produced from reacting $18.48g$ of barium chloride with an excess amount of sodium sulfate?

Answer 1

We have to know that the reaction which is carried out between barium chloride and sodium sulphate is a type of double displacement reaction as the ions for both the reactants are displacing to give new products. It can sometimes be called a double replacement reaction as the ions are replaced with each other.

Complete answer:
We have to know that the double displacement reaction between barium chloride and sodium chloride can be represented as:
$BaC{l_2}(aq) + N{a_2}S{O_4}(aq) \to BaS{O_4}(s) \downarrow + 2NaCl(aq)$
We have to remember that the barium sulphate is formed as a white precipitate in this reaction. Now we are asked to give the amount of barium sulphate formed when 18.48 grams of barium chloride reacts with excess amount of sodium sulfate this can be solved as shown below:
Moles of Barium chloride=$\dfrac{{18.48g}}{{208.23g/mol}}$
$= 8.87 \times {10^{2 - }}mol$
Barium chloride is the limiting reagent.
At most, we can get $8.87 \times {10^{2 - }}mol$ of $BaS{O_4}$ (s), so we can say that:
$= 8.87 \times {10^{ - 2}}mol \times 233.38g \cdot mo{l^{ - 1}}$
$= 2070{\text{ x }}{10^{ - 2}}g$
Barium sulfate is as soluble in water as a brick, which is why we used sulfate ion to precipitate the salt.
So this is how we can calculate the amount of barium sulphate.

Note:
We need to remember that the number of moles can be the given mass divided by the molar mass for any substance. If the mass is given for any substance it is easy to calculate the number of moles, as molecular mass can be calculated by itself. Barium sulphate so formed is a white precipitate which cannot be soluble in water, thus it stays in a precipitated form.