Đáp án + Giải thích các bước giải:

$a)PTHH:$

$Fe+2HCl\to FeCl_2+H_2\uparrow$

$Fe_2O_3+6HCl\to 2FeCl_3+3H_2O$

$FeCl_2+2AgNO_3\to Fe(NO_3)_2+2AgCl\downarrow$

$FeCl_3+3AgNO_3\to Fe(NO_3)_3+3AgCl\downarrow$

$Fe(NO_3)_2+2NaOH\to Fe(OH)_2\downarrow+2NaNO_3$

$Fe(NO_3)_3+3NaOH\to Fe(OH)_3\downarrow+3NaNO_3$

$b)n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol);n_{AgCl}=\dfrac{86,1}{143,5}=0,6(mol)$

Theo các PT: $n_{Fe}=n_{FeCl_2}=n_{H_2}=0,1(mol)$

$\Rightarrow m_{Fe}=0,1.56=5,6(g)$

$\Rightarrow n_{FeCl_3}=\dfrac{1}{3}.(n_{AgCl}-2.n_{FeCl_2})=\dfrac{1}{3}(0,6-0,2)=\dfrac{4}{3}(mol)$

$\Rightarrow n_{Fe_2O_3}=\dfrac{1}{2}n_{FeCl_3}=\dfrac{2}{3}(mol)$

$\Rightarrow m_{Fe_2O_3}=\dfrac{2}{3}.160=106,67(g)$

$\Rightarrow \%m_{Fe_2O_3}=\dfrac{106,67}{106,67+5,6}.100\%=95,11\%$

$\Rightarrow \%m_{Fe}=100-95,11=4,89\%$

$c)$ Theo các PT: $n_{NaOH}=2n_{FeCl_2}+3n_{FeCl_3}=n_{AgCl}=0,6(mol)$

Đổi $400ml=0,4l\Rightarrow C_{M_{NaOH}}=\dfrac{0,6}{0,4}=1,5M$

$d)$ Theo các PT: $n_{Fe(OH)_2}=0,5n_{FeCl_2}=0,1(mol)$

$n_{Fe(OH)_3}=n_{FeCl_3}=\dfrac{4}{3}(mol)$

$\Rightarrow m_{\downarrow}=m_{Fe(OH)_2}+m_{Fe(OH)_3}=0,1.90+\dfrac{4}{3}.107=151,67(g)$

Vậy $m=151,67$