Question

The maximum amount of $Ba{SO}_4$ precipitated on mixing $20mL$ of $0.5M$ $Ba{Cl}_2$ with $20mL$ of $1M$ $H_2{SO}_4$ is?

• A. $0.25$ mole
• B. $0.5$ mole
• C. $1$ mole
• D. $0.01$ mole

Answer 1

The correct option is D $0.01$ mole

Given molarity of $BaC{l}_2$$=0.5M$

Given volume of $BaC{l}_2$$=20ml$

Number of moles of $BaC{l}_2$$=\frac{0.5}{1000}\times 20=0.01$ Moles

Given molarity of $H_{2}S{O}_4$$=1M$

Given volume of $H_{2}S{O}_4$$=20ml$

Number of moles of $H_{2}S{O}_4$$=\frac{1}{1000}\times 20=0.02$ Moles

Reaction of $BaC{l}_2$and $H_{2}S{O}_4$ :

$BaC{l}_2+H_{2}S{O}_4\to BaS{O}_4(\downarrow )+2HCl$

In the reaction $1$ mole of $BaC{l}_2$ reacts with $1$ mole of $H_{2}S{O}_4$ to give $1$ mole of $BaS{O}_4$ and $2$ moles of $HCl$.

Which implies that from the above given conditions the given $0.01$moles of $BaC{l}_2$ will react only with $0.01$moles of $H_{2}S{O}_4$.

Therefore $BaC{l}_2$ is the limiting reagent.

Again from the reaction $0.01$moles of $BaC{l}_2$ will react with $0.01$moles of $H_{2}S{O}_4$ to give $0.01$ moles of $BaS{O}_4$ and $0.02$ moles of $HCl$.

Therefore, the correct option is D i.e $0.01$ moles of $BaS{O}_4$ .