Đáp án:
Giải thích các bước giải:
`a.2Al+6HCl→2AlCl_3+3H_2↑`
`Al_2O_3+6HCl→2AlCl_3+3H_2O`
`2Cu+O_2→2CuO`
`n_{CuO}=\frac{12}{80}=0,15(mol)`
`n_{Cu}=n_{CuO}=0,15(mol)`
`b.m_{Cu}=0,15.64=9,6(g)`
`n_{H_2}=\frac{6,72}{22,4}=0,3(mol)`
`n_{Al}=2/3.n_{H_2}=0,2(mol)`
`m_{Al}=0,2.27=5,4(g)`
`m_{Al_2O_3}=25,2-5,4-9,6=10,2(g)`
`%m_{Cu}=\frac{9,6}{25,2}.100=38,095%`
`%m_{Al}=\frac{5,4}{25,2}.100=21,429%`
`%m_{Al_2O_3}=\frac{10,2}{25,2}.100=40,476%`
`#Devil`
