Question

To $50L$ of $0.2NNaOH,2.5L$ of $2NHCl$ and $15L$ of $0.1NFeC{l}_3$ solutions are added. What weight of $F{e}_2{O}_3$ can be obtained from the precipitate? Also report the normality of NaOH left in the resultant solution.

• A. $40g,0.05N$
• B. $20g,0.1N$
• C. $40g,0.1N$
• D. $20g,0.05N$

Answer 1

The correct option is A $40g,0.05N$

Equivalents of $NaOH=50\times 0.2=10$
Equivalents of $HCl=2.5\times 2=5$
$\text{Equivalents of}NaOH\text{left after reaction with}HCl=10-5=5$

$FeC{l}_3+3NaOH\to Fe(OH{)}_3\downarrow +3NaCl$
$2Fe(OH{)}_3\stackrel{\Delta }{\to }F{e}_2{O}_3+3H_{2}O$

$FeC{l}_3$ reacts with $NaOH$ to give $Fe(OH{)}_3$ which on ignition gives $F{e}_2{O}_3$ .

$\therefore \text{Equivalents of}NaOH\text{used for}FeC{l}_3=\text{Equivalents of}Fe(OH{)}_3=\text{Equivalents of}F{e}_2{O}_3=15\times 0.1=1.5$
$\therefore \text{Equivalents of}NaOH\text{left finally}=5-1.5=3.5$
$\text{Total volume}=(50+2.5+15)L=67.5L$
$\text{Normality of}NaOH\text{in the resultant solution}=\frac{3.5}{67.5}\approx 0.05N$
$\frac{W_{F{e}_2{O}_3}}{M_{F{e}_2{O}_3}}\times 6=1.5\text{(n-factor for}F{e}_2{O}_3=6)$
$W_{F{e}_2{O}_3}=\frac{1.5\times 160}{6}=40g$