Đáp án:

Giải thích các bước giải:

`A:Mg:x(mol);Al:y(mol)`

`24x+27y=12,9(1)`

BTKL: `n_{Cl_2}={41,3-12,9}/{71}=0,4(mol)`

`Mg+Cl_2` $\xrightarrow{t^o}$ `MgCl_2`

`2Al+3Cl_2` $\xrightarrow{t^o}$ `2AlCl_3`

`2Al+6HCl->2AlCl_3+3H_2`

`Mg+2HCl->MgCl_2+H_2`

`CuO+H_2` $\xrightarrow{t^o}$ `Cu+H_2O`

BTKL: `n_{O\ pu}={20-16,8}/{16}=0,2(mol)`

Theo pt:

`n_{H_2\ pu}=n_{O\ pu}=0,2(mol)`

`->n_{H_2\ bd}={0,2}/{80\%}=0,25(mol)`

Theo pt:

`n_{H_2}+n_{Cl_2}=x+1,5y=0,65(2)`

`(1)(2)->x=0,2;y=0,3`

`->n_{Mg}=0,2(mol);n_{Al}=0,3(mol)`