Question

Equivalent weight of $Fe{C}_2{O}_4$ in the change $Fe{C}_2{O}_4\to F{e}^{3+}+C{O}_2$ is?

• A. $\frac{M}{3}$
• B. $\frac{M}{6}$
• C. $\frac{M}{2}$
• D. $\frac{M}{1}$

Answer 1

Answer: (A)

$\frac{M}{3}$

$Fe{C}_2{O}_4⟶F{e}^{3+}+C{O}_2$

Ox. no. of $Fe$ in $Fe{C}_2{O}_4=F{e}^{2+}$

Ox. no. of $C$ in $Fe{C}_2{O}_4=+3$

Ox. no. of $C$ in $C{O}_2=+4$

Electron change per molecule in oxidation of $Fe=1$

Eectron change per molecule in oxidation of $C=1\times 2=2$

$\therefore$ Total electron change (n-factor)$=1+2=3$

$\therefore$ Equation of weight of $Fe{C}_2{O}_4=\frac{Molecularweight}{3}$