Question

$A{s}_2{S}_3+\underset{–}{X}N{O}_3^{-}+\underset{–}{Y}{H}^{+}\to 2AS{O}_4^{3-}+\underset{–}{Z}N{O}_2+\underset{–}{A}S+2H_{2}O$ Here X,Y,Z and A respectively are

• A. 10, 4, 10, 3
• B. 5, 4, 5, 3
• C. 4, 10, 3, 10
• D. None of these

Answer 1

The correct option is A 10, 4, 10, 3

This is a very special case of redox reaction. Note that the given reaction involves two oxidations and one reduction.
a. Two oxidation half reactions are
$i.A{s}^{3+}\to As{O}_4^{3-}ii.S^{2-}\to S$
Now balance and add the two equations to get
$2A{s}^{3+}+3S^{2-}+8H_{2}O\to 2As{O}_4^{3-}+3S+16H^{+}+10e^{-}$
b. Reduction half reaction: $N{O}_3^{-}\to N{O}_2$
Now balance the reduction half reaction and proceed as usual .
$N{O}_3^{-}+2H^{+}+1e^{-}\to N{O}_2+H_{2}O$
Adding the two reactions (i) and (ii), we get
$A{s}_2{S}_3+10N{O}_3^{-}+4H^{+}\to 2As{O}_4^{3-}+10N{O}_2+3S+2H_{2}O$
Note : In the above type of cases, take total change in oxidation state (increase or decrease) for cross multiplication while adding oxidation and reductions. For example, in the given case, 10 electrons are lost per mole of $A{s}_2{S}_3$.