Solveeit Logo
Login

Question

Question: How would you balance the following equation: \(S\, + \) \(HN{O_3} \to \,{H_2}S{O_4} + \,N{O_2}\, ...

How would you balance the following equation:
S +S\, + HNO3→ H2SO4+ NO2 +H2OHN{O_3} \to \,{H_2}S{O_4} + \,N{O_2}\, + {H_2}O

Explanation

Solution

We will balance this chemical reaction by Oxidation number method
We all know that HNO3HN{O_3} is an oxidizing agent, it has oxidized Sulphur from oxidation state of0 in S to+6 in H2SO40\,in\,S\,to + 6\,in\,{H_2}S{O_4} . The oxidation number of Nitrogen has changed from+5 in HNO3 to +4 in NO2 + 5\,in\,HN{O_3}\,to\, + 4\,in\,N{O_2}. After balancing the oxidation numbers of the atoms at both sides, we need to balance the remaining atoms O and H by hit and trial method.

Complete answer:
We will balance this chemical reaction by Oxidation number method.
The balancing is done in following steps:
S +S\, + HNO3→ H2SO4+ NO2 +H2OHN{O_3} \to \,{H_2}S{O_4} + \,N{O_2}\, + {H_2}O

  1. Write the oxidation number of each atom in the equation
    SO + H+1N+5O−23→{S^O}\, + \,{H^{ + 1}}{N^{ + 5}}{O^{ - 2}}_3 \to H+12S+6O−24 + N+4O−22 + H+12O−2{H^{ + 1}}_2{S^{ + 6}}{O^{ - 2}}_4\, + \,{N^{ + 4}}{O^{ - 2}}_2\, + \,{H^{ + 1}}_2{O^{ - 2}}
  2. Identify the atoms which undergo change in oxidation number.S0 + HNO3  → H2S+6O4 +NO2 +H20{S^0}\, + \,HN{O_{3\,}}\, \to \,{H_2}{S^{ + 6}}{O_4}\, + N{O_2}\, + {H_2}0
    S + HN+5O3 → H2SO4 + N+4O2+H20S\, + \,H{N^{ + 5}}{O_3}\, \to \,{H_2}S{O_4}\, + \,{N^{ + 4}}{O_2} + {H_2}0
  3. Calculate the increase and decrease in oxidation number with respect to reactant atoms.S0 +HN+5O3 → H2S+6O4 + N+4O2 + H2O{S^0}\, + H{N^{ + 5}}{O_3}\, \to \,{H_2}{S^{ + 6}}{O_4}\, + \,{N^{ + 4}}{O_2}\, + \,{H_2}O
    Increase in oxidation number of Sulphur is from0 in S to +6 in H2SO40\,in\,S\,to\, + 6\,in\,{H_2}S{O_4}
    Decrease in oxidation number of Nitrogen is from+5 in HNO3 to +4 in NO2 + 5\,in\,HN{O_3}\,to\, + 4\,in\,N{O_2}
  4. Equate the increase and decrease in oxidation number.
    S + 6 HNO3→ H2SO4 +6NO2 +H2OS\, + \,6\,HN{O_3} \to \,{H_2}S{O_4}\, + 6N{O_2}\, + {H_2}O
    We can clearly see that N and S atoms are already balanced in the chemical reaction. Now, our job is to 5) balance the number of H and O atoms by hit and trial method.
    S + 6HNO3→ H2SO4 +6NO2 +2H2OS\, + \,6HN{O_3} \to \,{H_2}S{O_4}\, + 6N{O_2}\, + 2{H_2}O
    This is the required balanced equation.

Note: The balancing of a chemical reaction is done to equalize the atoms of different elements or compounds which are involved in it. It is done to fulfil the requirement of the law of conservation of mass. We need to identify carefully the oxidation number of the reactants and products. Also identify carefully the elements undergoing increase and decrease in oxidation number.