Question

In the redox reaction given below:

$xN{O}_3^{-}+yA{s}_2{S}_3+z{H}_{2}O\to As{O}_4^{3-}+NO+S{O}_4^{2-}+H^{+}$

What is the value of $\frac{x}{z}$?

Answer 1

$n$-factor of $A{s}_2{S}_3=3\times (6-(-2\left)\right)+2(5-3)=24+2\times 2=28$.

$n$-factor of $N{O}_3^{-}=5-2=3$.

The balanced reaction is as follows:

$28N{O}_3^{-}+3A{s}_2{S}_3+4H_{2}O\to 6As{O}_4^{3-}+28NO+9S{O}_4^{2-}+8H^{+}$

So, the value of $\frac{x}{z}=\frac{28}{4}=7$.