Question

Excess milliequivalents of $Br{O}_3^{⊝}$ in reaction (ii) is :

• A. $\frac{1}{6}$
• B. $\frac{11}{6}$
• C. $\frac{1}{36}$
• D. $\frac{11}{36}$

Answer 1

Answer: (A)

$\frac{1}{6}$

Total millimoles $Br{O}_3^{⊝}=20\times \frac{1}{60}=\frac{1}{3}$

Total millimoles of $As{O}_2^{⊝}=\frac{1}{60}\times 5=\frac{1}{12}$

According to the balanced equation for first and second reaction,
$\underset{5mmol}{5Se{O}_3^{2-}}+\underset{2mmol}{2Br{O}_3^{⊝}}+2H^{⨁}\to 5Se{O}_4^{2-}+B{r}_2+H_{2}O$ ....(i)

$\underset{1mmol}{Br{O}_3^{⊝}}+\underset{3mmol}{3As{O}_2^{⊝}}+H_{2}O\to B{r}^{⊝}+3As{O}_4^{3-}+6H^{⨁}$ .....(ii)

$\because 3$ mmol of $As{O}_2^{⊝}=1$ mmol of $Br{O}_3^{⊝}$ in equation (ii)

$\frac{1}{12}$ mmol of $As{O}_2^{⊝}=\frac{1}{3}\times \frac{1}{12}$ $=\frac{1}{36}$ mmol of excess $Br{O}_3^{⊝}$

Millimoles of $Br{O}_3^{⊝}$ reacted with $Se{O}_3^{2-}$ in equation (i) $=\frac{1}{36}$

$[Br{O}_3^{⊝}\equiv As{O}_2^{⊝}]$ from equation (ii)

$\frac{1}{36}$ mmol $=\frac{1}{12}$ mmol

$\frac{1}{36}\times 6$ mEq. $=\frac{1}{12}\times 2$ mEq.

Excess mEq. of $B{r}_3^{⊝}=\frac{1}{6}$ mEq.

Hence, answer $=\frac{1}{6}$ mEq.