${\mathrm{BrO}}_3^{\ominus }+5{\mathrm{Br}}^{\ominus }⟶{\mathrm{Br}}_2+3{\mathrm{H}}_2\mathrm{O}$

If 50 mL 0.1 M ${\mathrm{BrO}}_3^{\ominus }$ is mixed with 30 mL of 0.5 M ${\mathrm{Br}}_{ }^{\ominus }$solution that contains excess of ${\mathrm{H}}^{\oplus }$ ions, the moles of Br₂ formed are

Balance the equation: 

$$\begin{gathered} {\mathrm{BrO}}_3^{\ominus } + 5{\mathrm{Br}}^{\ominus } \stackrel{ }{\to } 3{\mathrm{Br}}_2+3{\mathrm{H}}_2\mathrm{O} \\ 1 \mathrm{mmol} 5 \mathrm{mmol} 1 \mathrm{mmol} \\ \mathrm{Given}: 50 \mathrm{x} 0.1 30 \mathrm{x} 0.5 \\ = 5 \mathrm{mmol} =15 \mathrm{mmol} \end{gathered}$$

$B{r}^{⊝}$ is the limiting reagent.

5 mmol of $B{r}^{⊝}$ gives $\Rightarrow$3 mmol of Br₂

15 mmol of $B{r}^{⊝}$ gives $\Rightarrow$$\frac{3\times 15}{5}=9 \mathrm{mmol}$ = 9 x 10^-3 mol