Question

$n$-factor for the following reaction is :
$Fe{S}_2\to F{e}_2{O}_3+S{O}_2$

• A. $8$
• B. $9$
• C. $10$
• D. $11$

Answer 1

The correct option is D $11$

$\stackrel{+2}{Fe}\stackrel{-1\times 2}{S_2}\to \stackrel{+3\times 2}{F{e}_2}\stackrel{-2\times 3}{O_3}+\stackrel{+4-2\times 2}{S{O}_2}$
$\underset{2x=4}{2F{e}^{2+}}\stackrel{0}{\to }\underset{2x=6}{F{e}_2^{3+}}+2e^{-}]n=\frac{2}{2}=1$
$\underset{2x=-2}{S_2^{2-}}\to \underset{2x=8}{\underset{2x-8=0}{2S{O}_2}}+10e^{-}]n=10$
Total $n$ factor $=10+1=11$