\(\begin{array}{l}
1800ml=1,8l\\
n_{H_2}=\frac{1,8}{24}=0,075(mol)\\
n_{Fe_3O_4}=\frac{10,44}{232}=0,045(mol)\\
Fe_3O_4+4H_2\xrightarrow{t^o}3Fe+4H_2O\\
Theo\,PT:\,n_{Fe_3O_4(pu)}=\frac{1}{4}n_{H_2}=0,01875(mol)\\
\to \%m_{Fe_3O_4(khu)}=\frac{0,01875}{0,045}.100\%=41,67\%\\
Theo\,PT:\,n_{Fe}=3n_{Fe_3O_4(pu)}=0,05625(mol)\\
\to m_{cr\,spu}=0,05625.56+(0,045-0,01875).232=9,24(g)
\end{array}\)