$n_{Al}=\dfrac{5,4}{27}=0,2(mol)$
$n_{Fe_3O_4}=\dfrac{17,4}{232}=0,075(mol)$
$8Al+3Fe_3O_4\xrightarrow{{t^o}} 9Fe+4Al_2O_3$
$\to$ theo lí thuyết: phản ứng vừa đủ
Đặt $n_{Al\text{pứ}}=x(mol)$
$\to$ dư $0,2-x$ mol $Al$
$n_{Fe}=\dfrac{9}{8}n_{Al}=1,125x(mol)$
$n_{H_2}=\dfrac{5,376}{22,4}=0,24(mol)$
$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2$
$Fe+H_2SO_4\to FeSO_4+H_2$
$Al_2O_3+3H_2SO_4\to Al_2(SO_4)_3+3H_2O$
$Fe_3O_4+4H_2SO_4\to FeSO_4+Fe_2(SO_4)_3+4H_2O$
$\to 1,5(0,2-x)+1,125x=0,24$
$\to x=0,16$
Vậy $H=\dfrac{0,16.100}{0,2}=80\%$