answer is

2HNO_{3} + 3H_{2}SO_{3} " src="http://educog.com/res/sfu/batchelo/Gallery/rarrow.gif" align="middle"> 2NO + H_{2}O + 3H_{2}SO_{4}

because

in reactant side

the oxidation number of N=+5

S=+4and

but product side

N=+2

then the oxidation number of N is reduced.

hope my answer is help for you,.....see you soon again.........

in 1st reaction

oxidation number of cl is reduced 0 to -1

2nd reaction

oxidation number of N is reduced +5 to +2

3rd reaction

oxidation number of I is reduced +5 to -1

4th reaction

oxidation number of cI is reduced 0 to -1

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