Question

Consider the reaction,
$aFe{S}_2+b{O}_2\to cF{e}_2{O}_3+dS{O}_2$
Which of the following options are incorrect for the above reaction?

• A. N- factor of $Fe{S}_2$ is 11
• B. The ratio of moles of a : b is 4 : 11
• C. The ratio of moles of a : b is 11 : 4
• D. The ratio of moles of c : d is 1 : 4

Answer 1

The correct option is C The ratio of moles of a : b is 11 : 4

In the reaction the
Oxidation state of Fe changes from +2 to +3
and of sulphur from -1 to +4
So, n-factor for $Fe{S}_2$ will be:
(no. of atoms * change in O.S) for Fe + (no. of atoms * change in O.S) for S
= 1*(3-2) + 2*(5)
=11
n-factor for $O_2$ = 2*2 =4
Now to balance the reaction we need to croos multiply the n-factors:
So
$4Fe{S}_2+11O_2\to 2F{e}_2{O}_3+8S{O}_2$
giving a= 4, b= 11, c= 2 d = 8
hence option (c) is the incorrect answer.