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# Balance the given equation:\begin{align} & S+HN{{O}_{3}}\to {{H}_{2}}S{{O}_{4}}+N{{O}_{2}}+{{H}_{2}}O \\ & \\ \end{align}  Verified
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Hint: First we have to find out the oxidation number of all atoms other than oxygen and hydrogen. Identify the oxidizing and reducing agents. Balancing of oxygen and hydrogen is done by hit and trial method.

The equation is balanced by the Oxidation number method.
Step by step we can balance the equation. In the first step write the skeletal equation and write the oxidation number of each atom.
\begin{align} & \text{0 +1 +5 -2 +1 +6 -2 +4 -2 +1 -2} \\ & \text{S + H N }{{\text{O}}_{\text{3}}}\to \text{ }{{\text{H}}_{\text{2}}}\text{ S }{{\text{O}}_{\text{4}}}\text{+ N }{{\text{O}}_{\text{2}}}\text{ + }{{\text{H}}_{\text{2}}}\text{ O} \\ & \\ \end{align}
Next, find out the elements which change oxidation number.

The oxidation number of S increases from 0 to +6. And the oxidation number of N decreases from +5 to +4.
Next, find out the total change in oxidation number.
Since, there is only one sulfur in both L.H.S and R.H.S, the total increase in oxidation number is 6. The number of N atom on L.H.S and R.H.S is one, hence the total decrease in oxidation number is 1.
Next, balance the oxidation number.
Since the total increase in the oxidation number is 6 and the total decrease is 1, therefore multiply and by 6.
\begin{align} & S+6HN{{O}_{3}}\to {{H}_{2}}S{{O}_{4}}+6N{{O}_{2}}+{{H}_{2}}O \\ & \\ \end{align}
Next balance all atoms other than H and O.
No need to balance the equation because there is one S atom on both sides and six N atoms on both sides.
Now balance the oxygen and hydrogen by hit and trial method.
There are 18 oxygen atoms on the L.H.S and 17 oxygen atoms on the R.H.S, therefore, add one${{H}_{2}}O$ molecule to the R.H.S.
\begin{align} & S+6HN{{O}_{3}}\to {{H}_{2}}S{{O}_{4}}+6N{{O}_{2}}+2{{H}_{2}}O \\ & \\ \end{align}
The hydrogen atoms get balanced automatically.

Hence, the balanced equation is: $S+6HN{{O}_{3}}\to {{H}_{2}}S{{O}_{4}}+6N{{O}_{2}}+2{{H}_{2}}O$

Note: For balance, you can only use water molecules that don't use oxygen atoms to balance the equation. If the equation is in an acidic medium hydrogen ion is used to balance the hydrogen atoms. If the equation is in basic medium hydroxyl ion is used to balance the hydrogen atoms.