Question

Balance the following equation by oxidation number method.
$HgS+HCl+HN{O}_3\to H_2\left[HgC{l}_4\right]+NO+S+H_{2}O$

Answer 1

Writing the oxidation numbers of the atoms.
$\stackrel{+2}{Hg}\stackrel{-2}{S}+\stackrel{+1}{H}\stackrel{-1}{Cl}+\stackrel{+1}{H}\stackrel{+5}{N}\stackrel{-2}{O_3}\to \stackrel{+1}{H_2}\stackrel{+2}{Hg}\stackrel{-1}{C{l}_4}+\stackrel{+2}{N}\stackrel{-2}{O}+\stackrel{0}{S}+\stackrel{+1}{H_2}\stackrel{-2}{O}$
The oxidation numbers of $S$ and $N$ have changed.
$Hg\stackrel{-2}{S}\to \stackrel{0}{S}$ .........(i)
$H\stackrel{+5}{N}{O}_3\to \stackrel{+2}{N}O$ ........(ii)
Increase in Ox. no. of $S$$=2$ units per HgS molecule
Decrease in Ox. no. of $N$$=3$ units per $HN{O}_3$ molecule
Multiplying eq. (i) by $3$ and eq. (ii) by $2$ as to make increase and decrease equal
$3HgS+2HN{O}_3\to 3S+2NO$
Balancing Hg and chlorine,
$3HgS+2HN{O}_3+12HCl\to 3H_{2}HgC{l}_4+3S+2NO$
To balance hydrogen and oxygen, $4H_{2}O$ molecules are added on RHS. Hence, the balanced equation is
$3HgS+2HN{O}_3+12HCl\to 3H_{2}HgC{l}_4+3S+2NO+4H_{2}O$.