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 #
The equivalent weight is183.82 g.
The usual formula for calculating the equivalent weight (
For a disproportionation reaction, as you have above, we must modify the formula to
#"EW" = "MW"/N = "MW"/n_1 + "MW"/n_2#
We can factor out
#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
For the reduction half-reaction
For the overall reaction
Here's another way to solve the problem.
The equivalent weight is the molecular weight (
#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.
We see that 2 mol of electrons are transferred in the reaction.
Since 3 mol of
The equivalent weight of