Question

$Br{O}_3^{\ominus }+5B{r}^{\ominus }\to B{r}_2+3H_{2}O$
If $50$ mL of $0.1$ M $Br{O}_3^{\ominus }$ is mixed with $30$ mL of $0.5MB{r}^{\ominus }$ solution that contains excess of $H^{\oplus }$ ions, the number of moles of $B{r}_2$ formed is:

• A. $6.0\times {10}^{-4}$
• B. $1.2\times {10}^{-4}$
• C. $9.0\times {10}^{-3}$
• D. $1.8\times {10}^{-3}$

Answer 1

Answer: (C)

$9.0\times {10}^{-3}$

Balance the equation:
$\underset{1mmol}{Br{O}_3^{\ominus }}+\underset{5mmol}{5B{r}^{\ominus }}\to \underset{1mmol}{3B{r}_2}+3H_{2}O$
Given: $\underset{=5mmol}{50\times 0.1}\underset{=15mmol}{30\times 0.5}$
$B{r}^{\ominus }$ is the limiting reagent.
$5$ mmol of $B{r}^{\ominus }$ gives $\Rightarrow$ $3$ mmol of $B{r}_2$
$14$ mmol of $B{r}^{\ominus }$ gives $\Rightarrow \frac{3\times 15}{5}=9mmol$
$=9\times {10}^{-3}mol$