Question

In a balanced equation, $H_{2}S{O}_4+xHI\to H_{2}S+y{I}_2+z{H}_{2}O$, the values of $x,y$ and $z$ are respectively :

• A. $x=3,y=5,z=2$
• B. $x=4,y=8,z=5$
• C. $x=8,y=4,z=4$
• D. $x=5,y=3,z=1$

Answer 1

Answer: (C)

$x=8,y=4,z=4$

To balance the number of iodine atoms, $HI$ is multiplied with $2$.
Iodine is oxidized. Its oxidation state changes from $-1$ to $0$. Net increase is $+2$.
Sulphur is reduced. Its oxidation state changes from $6$ to $-2$. Net decrease is $8$.
To balance the net increase and net decrease in the oxidation number, factor $4$ is used. Thus, coefficients of $HI$ and $I_2$ are multiplied by $4$.
To balance oxygen atoms, four water molecules are required on the right.
Thus, $x=8$, $y=4$ and $z=4$.