# According to the balanced equation below, how many moles of ClO^-2 (aq) are needed to react completely with 20. mL of 0.20 M KMnO_4 solution?

## $2 {H}_{2} O \left(l\right) + 4 M n {O}_{4} \left(a q\right) + 3 C l {O}_{2}^{-} \left(a q\right) \to 4 M n {O}_{2} \left(s\right) + 3 C l {O}_{4}^{-} \left(a q\right) + 4 O {H}^{-} \left(a q\right)$

Dec 18, 2016

$3 \cdot m m o l$ $\text{chlorite anion}$ are required.

#### Explanation:

$\text{Reduction}$
$M n {O}_{4}^{-} + 4 {H}^{+} + 3 {e}^{-} \rightarrow M n {O}_{2} \left(s\right) + 2 {H}_{2} O \left(l\right)$ $\left(i\right)$

$\text{Oxidation}$
$C l {O}_{2}^{-} + 2 {H}_{2} O \rightarrow C l {O}_{4}^{-} + 4 {H}^{+} + 4 {e}^{-}$ $\left(i i\right)$

$\text{Overall}$ $4 \times \left(i\right) + 3 \times \left(i i\right)$
$4 M n {O}_{4}^{-} + 3 C l {O}_{2}^{-} + 4 {H}^{+} \rightarrow 4 M n {O}_{2} \left(s\right) + 3 C l {O}_{4}^{-} + 2 {H}_{2} O \left(l\right)$

$\text{Moles of } K M n {O}_{4} = 20 \times {10}^{-} 3 L \times 0.20 \cdot m o l \cdot {L}^{-} 1 = 4.0 \times {10}^{-} 3 \cdot m o l$

Given the equation, we need 3/4 equiv of chlorite anion,

$\frac{3}{4} \times 4.0 \times {10}^{-} 3 \cdot m o l$

What would you observe in this reaction? What was the macroscopic colour change?