\(PTHH:2Al+Fe_2O_3\rightarrow Al_2O_3+2Fe\)
Ban đầu : 1,8_____0,6________________
Phứng :1,2________0,6_____________
Sau phứng : 0,6__0_________0,6_______0
\(n_{Fe2O3}=\frac{96}{160}=0,6\left(mol\right)\)
\(n_{Al}=\frac{48,6}{27}=1,8\)
\(2Al+2H_2O+2NaOH\rightarrow2NaAlO_2+3H_2\)
0,6_______________________________0,9
\(\rightarrow V_{H2}=0,9.22,4=20,16\left(l\right)\)
Vậy chọn D