nMg=\(\frac{2.4}{24}\)=0.1(mol)
PTPU: Mg+ 2HCl -----> MgCl2 +H2
0.1 0.1 (mol)
Vh2=0.1*22.4=2.24(l)
nFe3O4=\(\frac{11.6}{232}\)=0.05(mol)
Fe3O4 +4H2 ➞3Fe+4H20
0.25 0.1 0.075(mol)
( Xét \(\frac{0.05}{1}\) >\(\frac{0.1}{4}\)➞Fe3O4 dư , tính theo chất pứ hết)
mfe(sau pứ)=0.075*56=4.2(g)
good luck!!!