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

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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.

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