Warning! This is a long answer. The balanced equation is
"3KClO"_3 + "H"_2"SO"_4 → "ClO"_4^(-) + 2"ClO"_2 + "SO"_4^(2-) + 3"K"^+ + "H"_2"O"
We start with the unbalanced equation:
"KClO"_3 + "H"_2"SO"_4 → "ClO"_4^(-) + "ClO"_2 + "SO"_4^(2-) + "K"^+ + "H"_2"O"
Step 1. Identify the atoms that change oxidation number
stackrel(color(blue)(+1))("K") stackrel(color(blue)(+5))("Cl") stackrel(color(blue)(-2))("O")_3 + stackrel(color(blue)(+1))("H")_2 stackrel(color(blue)(+6))("S") stackrel(color(blue)(-2))("O")_4 → (stackrel(color(blue)(+7))("Cl") stackrel(color(blue)(-2))("O")_4)^(-) + stackrel(color(blue)(+4))("Cl") stackrel(color(blue)(-2))("O")_2 + (stackrel(color(blue)(+6))("S") stackrel(color(blue)(-2))("O")_4)^(2-) + (stackrel(color(blue)(+1))("K"))^+ + stackrel(color(blue)(+1))("H")_2 stackrel(color(blue)(-2))("O")
Left hand side: "K" = +1; "Cl" = +5; "O" = -2; "H" = +1; "S" = +6
Right hand side: "Cl in ClO"_4^(-) = +7; "O" = -2; "Cl in ClO"_2 = +4; "S" = +6; "K" = +1; "H" = +1
The changes in oxidation number are:
"Cl": +5 → +7 in "ClO"_4^(-); Change =+2
"Cl": +5 → +4 in "ClO"_2; Change = -1
Step 2. Equalize the changes in oxidation number
We need 2 atoms of "Cl" in "ClO"_2 for every 1 atom of "Cl" in "ClO"_4^-. This gives us total changes of -2 and +2.
Step 3. Insert coefficients to get these numbers
color(red)(3)"KClO"_3 + "H"_2"SO"_4 → color(red)(1)"ClO"_4^(-) + color(red)(2)"ClO"_2 + "SO"_4^(2-) + "K"^+ + "H"_2"O"
Step 4. Balance "K" by adding "K"^+ ions to the appropriate side
color(red)(3)"KClO"_3 + "H"_2"SO"_4 → color(red)(1)"ClO"_4^(-) + color(red)(2)"ClO"_2 + "SO"_4^(2-) + color(red)(3)"K"^+ + "H"_2"O"
Step 5. Balance charge
We have no net charge on the left or on the right, so there must be only 1 "SO"_4^(2-)
color(red)(3)"KClO"_3 + "H"_2"SO"_4 → color(red)(1)"ClO"_4^(-) + color(red)(2)"ClO"_2 + color(green)(1)"SO"_4^(2-) + color(red)(3)"K"^+ + "H"_2"O"
**Step 6. Balance "S" **
We have 1 "S" atom on the right, so we must have 1 "S" atom on the left
color(red)(3)"KClO"_3 + color(purple)(1)"H"_2"SO"_4 → color(red)(1)"ClO"_4^(-) + color(red)(2)"ClO"_2 + color(green)(1)"SO"_4^(2-) + color(red)(3)"K"^+ + "H"_2"O"
Step 7. Balance "H"
We have 2 "H" on the left, so we need 2 "H" on the right.
color(red)(3)"KClO"_3 + color(teal)(1)"H"_2"SO"_4 → color(red)(1)"ClO"4^-) + color(red)(2)"ClO"_2 + color(green)(1)"SO"_4^(2-) + color(red)(3)"K"^+ + color(orange)(2)"H"_2"O"
The balanced equation is
color(red)("3KClO"_3 + "H"_2"SO"_4 → "ClO"_4^(-) + 2"ClO"_2 + "SO"_4^(2-) + 3"K"^+ + "H"_2"O")
