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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.

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

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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}$$=0.05N
Total volume =70litre
Eq.of $Fe_2O_3=1.5$
$\large\frac{W}{M/6}$$=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
 

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