Question

To $50$ L of $0.2NNaOH$, $5$ L of $1NHCl$ and $15$ L of $0.1NFe{Cl}_3$ solutions are added. The mass (in g) of ${Fe}_2{O}_3$ that can be obtained from the precipitate is :

Answer 1

Equivalents of $NaOH=50\times 0.2=10$
Equivalents of $HCl=5\times 1=5$
Equivalents of $NaOH$ left after reaction with $HCl=10-5=5$

Also, $NaOH$ reacts with $Fe{Cl}_3$ to give $Fe{\left(OH\right)}_3$ which on ignition gives ${Fe}_2{O}_3$.

Equivalents of $NaOH$ used for $Fe{Cl}_3=$ Equivalents of $Fe{\left(OH\right)}_3$ $=$ Equivalents of ${Fe}_2{O}_3$ $=15\times 0.1=1.5$

Eq. of $NaOH$ left finally $=5-1.5=3.5$
$N_{NaOH}$ left $=\frac{3.5}{70}=0.05N$

Total volume of solution $=70$ L
Also, eq. of ${Fe}_2{O}_3=1.5$
$\frac{W}{\frac{M}{6}}=1.5$
$W_{{Fe}_2{O}_3}=\frac{1.5\times 160}{6}=40$ g