Balance the equation
`As_(2)S_(3) + NO_(3)^(-) to AsO_(4)^(3-) + NO + S`
in acidic medium

Explanation

This is very special case of redox reaction. Note that the given reaction involves two oxidations and one reduction.
a. Two oxidation half reactions are
i. `As^(3+)rarrAsO_(4)^(3-)`
ii. `S^(2-)rarrS`
First balancing these two half reductions:
i. `2As^(3+)+8H_(2)OrarrAsO_(4)^(3-)+16H^(o+)+4e^(-)`
ii. `3S^(2-)rarr3S+6e^(-)`
Now add the two equation to get
`2As^(3+)+3S^(2-)+8H_(2)Orarr2AsO_(4)^(3-)+3S+16H^(o+)+10e^(-)`....(i)
b. Reduction half reaction: `NO_(3)^(ө)rarrNO_(2)`
Now balance the reduction half reaction and proceed as usual.
`NO_(3)^(ө)+2H^(o+)+1e^(-)rarrNO_(2)+H_(2)O`....(ii)
Adding the two reactions (i) and (ii), we get
`As_(2)S_(3)+10NO_(3)^(ө)+4H^(o+)rarr2AsO_(4)^(3-)+10NO_(2)+3S+3H_(2)O`
Note: In the above types of cases, take total change in oxidation number (increase or decrese) for cross multiplication while adding oxidation and reduction half reactions. For example, in the given case `10` electrons are lost per mole of `As_(2)S_(3)`.