Đáp án:
Giải thích các bước giải:
`m_{Al_2O_3}=10.68%=6,8(\text{tấn})=6800(kg)=6800000(g)`
`n_{Al_2O_3}=\frac{6800000}{102}=200000/3(mol)`
`n_{Al_2O_3pư}=200000/3 .70%=140000/3(mol)`
$2Al_2O_3\xrightarrow{đp}4Al+3O_2↑$
`n_{Al}=2n_{Al_2O_3}=280000/3(mol)`
`m_{Al}=280000/3 .27=2520000(g)=2520(kg)=2,52(\text{tấn})`
`#Devil`