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 #

2 Answers
Jun 10, 2018

Answer:

The equivalent weight is183.82 g.

Explanation:

The usual formula for calculating the equivalent weight (#"EW"#) of a substance in a redox equation is

#"EW" = "MW"/n#

where

#"MW ="# the molecular weight of the substance and

#n color(white)(ml)=# the change in oxidation number

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

#"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 #"MW"# to get

#1/N = 1/n_1 + 1/n_2#

We can solve this equation to get

#color(blue)(bar(ul(|color(white)(a/a)N = (n_1n_2)/(n_1 + n_2)color(white)(a/a)|)))" "#

For the oxidation half-reaction

#"K"stackrelcolor(blue)("+5")("Cl")"O"_3 → "H"stackrelcolor(blue)("+7")("Cl")"O"_4#

#n = 2#

For the reduction half-reaction

#"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#

#"EW" = "MW"/(2/3) = "122.55 g"× 3/2= "183.82 g"#

Jun 11, 2018

Answer:

Here's another way to solve the problem.

Explanation:

The equivalent weight is the molecular weight (#"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: #"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: #color(white)(mmm)3"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 #"KClO"_3# are involved, the number of electrons transferred per mole of #"KClO"_3# is

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

#"Equivalent weight" = "MM"/n = "122.55 g"/(2/3) = "183.82 g"#

The equivalent weight of #"KClO"_3# in this reaction is 183.82 g.