Question

n-factor for the following reaction is :
${\mathrm{FeS}}_2\to {\mathrm{Fe}}_2{\mathrm{O}}_3+{\mathrm{SO}}_2$

• A. 8
• B. 9
• C. 10
• D. 11

Answer 1

The correct option is D

11

The correct option is (D):-

Explanation for the correct option:

Step 1: The reactions involved :

$4{\mathrm{FeS}}_2+11{\mathrm{O}}_2\to 2{\mathrm{Fe}}_2{\mathrm{O}}_3+8{\mathrm{SO}}_2$

which is formed by two half reactions.

Step 2: Finding the n-factor in the reaction.

$\stackrel{+2-1\times 2}{{\mathrm{FeS}}_2}\to \stackrel{+3\times 2-2\times 3}{{\mathrm{Fe}}_2{\mathrm{O}}_3}+\stackrel{+4-2\times 2}{{\mathrm{SO}}_2}$

$\underset{2\mathrm{x}=4}{2{\mathrm{Fe}}^{2+}}\stackrel{0}{\to }\underset{2x=6}{F{e}_2^{3+}}+2e^{-}]n=\frac{2}{2}=1$

Further,

$\underset{2\mathrm{x}=-2}{{\mathrm{S}}_2^{2-}}\to \underset{\underset{2\mathrm{x}=8}{2\mathrm{x}-8=0}}{2{\mathrm{SO}}_2}+10{\mathrm{e}}^{-}]\mathrm{n}=10$

Hence, the total n-factor would be= 10+1 = 11

Therefore, the correct option is (D):- 11