A $1.0 ext{ g}$ sample of $ ext{Fe}_2 ext{O}_3$ solid of $55.2\%$ purity is dissolved in acid and

Question

A $1.0	ext{ g}$ sample of $	ext{Fe}_2	ext{O}_3$ solid of $55.2\%$ purity is dissolved in acid and reduced by heating the solution with zinc dust. The resultant solution is cooled and made up to $100.0	ext{ mL}$. An aliquot of $25.0	ext{ mL}$ of this solution requires $17.0	ext{ mL}$ of $0.0167	ext{ M}$ solution of an oxidant for titration. Calculate the number of electrons taken up by one molecule of the oxidant in the reaction of the above titration. (Given: Molar mass of $	ext{Fe}_2	ext{O}_3 = 160	ext{ g/mol}$)

Professor PC Thomas Classes · Answer

Mass of pure $	ext{Fe}_2	ext{O}_3$ in the sample $= 1.0 	imes 0.552 = 0.552	ext{ g}$. 
Moles of $	ext{Fe}_2	ext{O}_3 = 0.552 / 160 = 3.45 	imes 10^{-3}	ext{ mol} = 3.45	ext{ mmol}$. 
Moles of $	ext{Fe}$ atoms $= 2 	imes 3.45 = 6.9	ext{ mmol}$. 
Zinc dust reduces all $	ext{Fe}^{3+}$ to $	ext{Fe}^{2+}$. Thus, total $	ext{Fe}^{2+}$ in $100	ext{ mL} = 6.9	ext{ mmol}$. 
In a $25.0	ext{ mL}$ aliquot, moles of $	ext{Fe}^{2+} = 6.9 / 4 = 1.725	ext{ mmol}$. 
During the titration, the oxidant oxidizes $	ext{Fe}^{2+}$ back to $	ext{Fe}^{3+}$ (n-factor of Fe = 1). 
Equivalents of oxidant = Equivalents of $	ext{Fe}^{2+} = 1.725	ext{ meq}$. 
Let the n-factor (electrons taken up) of the oxidant be $n$. 
$17.0 	imes 0.0167 	imes n = 1.725 \implies 0.2839n = 1.725 \implies n = 1.725 / 0.2839 = 6.07 \approx 6$.