Question

The reaction given below is a balanced reaction where $a,b,c,d,e,f$ and $g$ are the stoichiometric coefficients.
$aHgS+bHCl+cHN{O}_3\to d{H}_{2}HgC{l}_4+eNO+fS+g{H}_{2}O$
Find $a+b+c+d+e+f+g$

Answer 1

Writing the chemical reaction given along with the oxidation number of the atoms involved:
$\stackrel{+2-2}{HgS}+\stackrel{+1-1}{HCl}+\stackrel{+1+5-2}{HN{O}_3}\to \stackrel{+1}{H_2}\stackrel{+2-1}{HgC{l}_4}+\stackrel{+2-2}{NO}+\stackrel{0}{S}+\stackrel{+1-2}{H_{2}O}$
The oxidation number of $S$ and $N$ have changed.
$Hg\stackrel{-2}{S}\to \stackrel{0}{S}\cdot \cdot \cdot \left(i\right)\stackrel{+5}{HN{O}_3}\to \stackrel{+2}{N}O\cdot \cdot \cdot \left(ii\right)$
Increase in oxidation number of $S$ = $2$ units per $HgS$ molecule
Decrease in Oxidation number 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
$3HgS+2HN{O}_3+12HCl\to 3H_{2}HgC{l}_4+3S+2NO+4H_{2}O$
So, $a+b+c+d+e+f+g=3+2+12+3+3+2+4=29$