Question

If $0.7$ moles of barium chloride is treated with $0.4$ mole of potassium sulphate, number of moles of barium sulphate formed is:

• A. $0.7$
• B. $0.4$
• C. $0.35$
• D. $0.2$

Answer 1

The correct option is B $0.4$

The reaction is $BaC{l}_2+K_{2}S{O}_4\to BaS{O}_4+2KCl$.

One mole of barium chloride reacts with one mole of potassium sulphate to form one mole of barium sulphate and two moles of potassium chloride.

Here, potassium sulphate is the limiting reagent. Hence, $0.4$ mol of potassium sulphate will react with $0.4$ mol of barium chloride to form $0.4$ mol of barium sulphate and $0.8$ mol of potassium chloride respectively.

Hence, option B is correct.