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")#
