Question

For the reaction $F{e}_2{S}_3+5O_2\to 2FeS{O}_4+S{O}_2$ the equivalent weight of $F{e}_2{S}_3$ is?

• A. $\frac{M}{4}$
• B. $\frac{M}{16}$
• C. $\frac{M}{22}$
• D. $\frac{M}{20}$

Answer 1

The correct option is C $\frac{M}{22}$

Molecular weight of $F{e}_2{S}_3$ $=55.845\times 2+32\times 3=307.9$ g

In $F{e}_2{S}_3$, the oxidation state of $S$ is -2

In $FeS{O}_4$, the oxidation state of $S$ is 6

In $S{O}_2$, the oxidation state of $S$ is +4

$n-$ factor = Total number of electrons lost or gained =$2\times 3+2\times 6+1\times 4=22$

$Equivalentweight=\frac{Molecularweight}{nfactor}=\frac{M}{22}$