Question

n-factor for ${\mathrm{H}}_2\mathrm{S}$ during its oxidation to ${\mathrm{SO}}_2$ is:

Answer 1

Step 1: Reaction of oxidation of ${\mathrm{H}}_2\mathrm{S}$ to ${\mathrm{SO}}_2$:

$2{\mathrm{H}}_2{\mathrm{S}}_{\left(\mathrm{g}\right)}+3{\mathrm{O}}_{2\left(\mathrm{g}\right)}\to 2{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}+2{\mathrm{SO}}_{2\left(\mathrm{g}\right)}$

Step 2: Finding the n-factor:

n-factor of ${\mathrm{H}}_2\mathrm{S}$:

$$\begin{gathered} 2\left(1\right)+\mathrm{n}=0 \\ \mathrm{n}=-2 \end{gathered}$$

n-factor of ${\mathrm{SO}}_2$:

$$\begin{gathered} \mathrm{n}+2(-2)=0 \\ \mathrm{n}=4 \end{gathered}$$

Therefore, the n-factor of ${\mathrm{H}}_2\mathrm{S}$ during its oxidation to ${\mathrm{SO}}_2$ is = $4-(-2)=6$