What will be the equivalent weight of KClO_3 in the following equation ?

3KClO_3 + 3H_2SO_4 → 3KHSO_4 + HClO_4 + 2ClO_2 + H_2O

Jun 10, 2018

The equivalent weight is183.82 g.

Explanation:

The usual formula for calculating the equivalent weight ($\text{EW}$) of a substance in a redox equation is

$\frac{\text{EW" = "MW}}{n}$

where

$\text{MW =}$ the molecular weight of the substance and

$n \textcolor{w h i t e}{m l} =$ the change in oxidation number

For a disproportionation reaction, as you have above, we must modify the formula to

$\frac{\text{EW" = "MW"/N = "MW"/n_1 + "MW}}{n} _ 2$

where

${n}_{1}$ and ${n}_{2}$ are the $n$-values for the two half-reactions and
$N$ is the equivalent $n$-value for the overall reaction

We can factor out $\text{MW}$ to get

$\frac{1}{N} = \frac{1}{n} _ 1 + \frac{1}{n} _ 2$

We can solve this equation to get

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} N = \frac{{n}_{1} {n}_{2}}{{n}_{1} + {n}_{2}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

For the oxidation half-reaction

${\text{K"stackrelcolor(blue)("+5")("Cl")"O"_3 → "H"stackrelcolor(blue)("+7")("Cl")"O}}_{4}$

$n = 2$

For the reduction half-reaction

${\text{K"stackrelcolor(blue)("+5")("Cl")"O"_3 → stackrelcolor(blue)("+4")("Cl")"O}}_{2}$

$n = 1$

For the overall reaction

N = (2 × 1)/(2 + 1) = 2/3

$\text{EW" = "MW"/(2/3) = "122.55 g"× 3/2= "183.82 g}$

Jun 11, 2018

Here's another way to solve the problem.

Explanation:

The equivalent weight is the molecular weight ($\text{MM}$) divided by the number of electrons transferred per mole of reactant ($n$).

color(blue)(bar(ul(|color(white)(a/a)"Equivalent weight" = "MM"/ncolor(white)(a/a)|)))" "

We need the moles of electrons, so let's balance the equation by the ion-electron method.

Unbalanced equation: ${\text{KClO"_3 → "KClO"_4 + "ClO}}_{2}$

Reduction: 1 × ["KClO"_3 + "H"_2"O" → "KClO"_4 + "2H"^"+" + 2"e"^"-"]
Oxidation: 2 × ul(["KClO"_3 + "2H"^"+" + "e"^"- "→ "K"^"+" + "ClO"_2 + "H"_2"O"]color(white)(mmm))
Overall: $\textcolor{w h i t e}{m m m} 3 \text{KClO"_3 + "2H"^"+" → "KClO"_4 + 2"K"^"+" + "2ClO"_2 + "H"_2"O}$

We see that 2 mol of electrons are transferred in the reaction.

Since 3 mol of ${\text{KClO}}_{3}$ are involved, the number of electrons transferred per mole of ${\text{KClO}}_{3}$ is

n = ("2 mol e"^"-")/("3 mol KClO"_3) = 2/3color(white)(l) "mol electrons per mole of KClO"_3"

$\text{Equivalent weight" = "MM"/n = "122.55 g"/(2/3) = "183.82 g}$

The equivalent weight of ${\text{KClO}}_{3}$ in this reaction is 183.82 g.