Balance the equation stepwise.
Write down the physical states of reactants and products in reactions.
`SO_(2) + 2H_(2) S to 3S +2H_(2)O`

Explanation

Step 1: Write the equation given.
`SO_(2)+H_(2) S to S+ H_(2)O…………(1)`
Count the number of atoms jof each element in products and reactants.

Step 2: It is seen from the above table, number of atoms on botyh sides are not the same. It means the equation (1) needs to be balanced.
Step 3: We will start with oxygen. On product side we need 2 atoms of oxygen, hence coefficient 2 for `H_(2)O.` Now the equation will be
`therefore SO_(2)+H_(2)S to S+ 2H_(2)O .............(2)`
Again count the number of atoms of elements on both the sides.

Still the equation is unbalanced. To balance sulphur and hydrogen atoms on reactant side, write coefficient 2 for `H_(2)S`
`therefore SO_(2)+2H_(2) S to S+2H_(2) O ..........(3)`
Again count the number of atoms on both the sides.

Now if we count number of atoms of hydrogen and oxygen, they are equal. But number of atoms of sulphur is not balanced. Now there are 3 atoms of sulphur on reactant side, whereas on product sides it is only one atom. Hence, we write coefficient 3 for sulphur on R.H.S. Now the equation becomes.
`SO_(2) +2H_(2)S to 3S + 2H_(2)O ............(4)`
Again count number, of atoms in equation (4)

It means equation (4) is a balanced equation.