Chat with tutor

Ask Questions, Get Answers


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.

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

1 Answer

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.
Help Clay6 to be free
Clay6 needs your help to survive. We have roughly 7 lakh students visiting us monthly. We want to keep our services free and improve with prompt help and advanced solutions by adding more teachers and infrastructure.

A small donation from you will help us reach that goal faster. Talk to your parents, teachers and school and spread the word about clay6. You can pay online or send a cheque.

Thanks for your support.
Please choose your payment mode to continue
Home Ask Homework Questions
Your payment for is successful.
Clay6 tutors use Telegram* chat app to help students with their questions and doubts.
Do you have the Telegram chat app installed?
Already installed Install now
*Telegram is a chat app like WhatsApp / Facebook Messenger / Skype.