$a)PTHH:8Al+3Fe_3O_4\xrightarrow{t^o}4Al_2O_3+9Fe$

$n_{Al_2O_3}=\dfrac{30,4}{102}=\dfrac{76}{255}(mol)$

Theo PT: $n_{Fe}=\dfrac{9}{4}n_{Al_2O_3}=\dfrac{57}{85}(mol)$

$\Rightarrow m_{Fe}=\dfrac{57}{85}.56\approx 37,55(g)$

$b)$ Theo PT: $n_{Al(p/ứ)}=2n_{Al_2O_3}=\dfrac{152}{255}(mol)$

$\Rightarrow m_{Al(dư)}=18,9-\dfrac{152}{255}.27\approx 2,8(g)$

$\Rightarrow n_{Al(dư)}=\dfrac{2,8}{27}=\dfrac{14}{135}(mol)$

Theo PT: $n_{Fe_3O_4(+Al dư)}=\dfrac{3}{8}n_{Al}=\dfrac{7}{180}(mol)$

$\Rightarrow m_{Fe_3O_4(+Al dư)}=\dfrac{7}{180}.232\approx 9,02(g)$

$c)PTHH:Fe+2HCl\to FeCl_2+H_2\uparrow$

Theo PT: $n_{H_2}=n_{Fe}=\dfrac{57}{85}(mol)$

$\Rightarrow m_{H_2}=\dfrac{57}{85}.2\approx 1,3(g)$

Và $m_{Fe}\approx 37,55(g)$

$\Rightarrow m_{\text{dd sau}}=37,55+100-1,3=136,25(g)$