How do you balance this equation: KMnO4 + H2SO3 -> K2SO4 + MnSO4 + H2SO4 + H20?

Question

I know that the answer is :

2KMnO4+5H2SO3->K2SO4+2MnSO4+2H2SO4+3H2...

but I don't know how to get that answer. Can someone please explain it to me? please and thank you.

Answer 1

Count the number of each element  on each side of the reaction.

You have to put numbers infront of each molecule (H2), KMnO4, etc etc...) to balance out the elements so they are equal on each side of the reaction. If you take the number from the answer and multiply each of the elements by it then they will be equal to the other side.

Left side has 2KMnO4 + 5H2SO3

This means 2 times 1 K element = 2 K elements

2 times 1 Mn = 2 Mn elements

2 times 4 O elements = 8 O elements

5 times 2 H = 10 H

5 times 1 S = 5 S

5 x 3 O = 15 O

So that makes the left side of the reaction (left side = reactants) have 2 K, 2 Mn, 5 S, 10 H and 23 O (15 O +8 O)

Now to the right side (products)

The first compound K2SO4 has no number in front of it meaning it needs to not be changed giving us

2 K, 1 S, 4 O

The next one is 2MnSO4

2 x 1 Mn = 2 Mn

2 x 1 S = 2 S

2 x 4 O = 8 O

Then 2 H2SO4

2 x 2 H = 4 H

2 x 1 S = 2 S

2 x 4 O = 8 O

The last one 3 H2O

3 x 2 H = 6 H

3 x 1 O = 3 O

Now lets add the right side together

2 K, 2 Mn, 5 S, 10 H, 23 O

If you scroll back up to the left side of the reactions totals these two will be equal.