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$\begin{align}

& S+HN{{O}_{3}}\to {{H}_{2}}S{{O}_{4}}+N{{O}_{2}}+{{H}_{2}}O \\

& \\

\end{align}$

Answer

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131.1k+ views

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.