Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

To 50litre of 0.2N NaOH ,5litre of 1N HCl and 15litre of 0.1N $FeCl_3$ solution are added .What weight of $Fe_2O_3$ can be obtained from the precipitate?Also report the normality of NaOH left in the resultant solution.

$\begin{array}{1 1}(a)\;0.05N,40g&(b)\;0.01N,30g\\(c)\;0.02N,35g&(d)\;0.03N,43g\end{array}$

Can you answer this question?

1 Answer

0 votes
Step 1:
Eq. of NaOH =$50\times 0.2=10$
Eq.of HCl=$5\times 1=5$
$\therefore$ Eq.of NaOH left after reaction with HCl=$10-5=5$
NaOH reacts with $FeCl_3$ 10 give $Fe(OH)_3$ which on ignition gives $Fe_2O_3$
$\therefore$ Eq.of NaOH used for $FeCl_3$=Eq of $Fe(OH)_3$=Eq.of $Fe_2O_3$
$\Rightarrow 15\times 0.1=1.5$
Step 2:
$\therefore$ Eq of NaOH left finally =5-1.5=3.5
$N_{NaOH}$ left =$\large\frac{3.5}{70}$$=0.05N
Total volume =70litre
Eq.of $Fe_2O_3=1.5$
$W_{FeO_3}=\large\frac{1.5\times 160}{6}$
$\Rightarrow 40g$
Hence (a) is the correct option.
answered Oct 31, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App