Chất rắn $B: Cu$ 

$PTHH : Cu +\dfrac{1}{2}O_2 \xrightarrow{to} CuO$ 

$n_{CuO}=\dfrac{2,75}{80}=0,034375(mol)$ 

theo pt 

$n_{Cu}=n_{CuO}=0,034375(mol)$ 

`->m_{Cu}=0,034375.64=2,2(g)`

$PTHH : 2Al + 6HCl \xrightarrow{} 2AlCl_3 +3H_2$ 

$Mg +2HCl \xrightarrow{} MgCl_2 +H_2$ 

$n_{H_2}=\dfrac{8,96}{22,4}=0,4(mol)$ 

đặt $x(mol)=n_{Al},y(mol)=n_{Mg}$ 

theo pthh ta có 

$1,5x+y=0,4(1)$ 

theo đề bài ta có 

$27x+24y=10-2,2=7,8(2)$ 

từ (1) và (2) suy ra 

$x=0,2(mol),y=0,1(mol)$ 

`->m_{Al}=0,2.27=5,4(g)`

`=> m_{Mg}=10-5,4-2,2=2,4(g)`

`Mg+2HCl->MgCl_2+H_2`

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

`2Cu+O_2` $\xrightarrow{t^o}$ `2CuO`

`->n_{Cu}=n_{CuO}={2,75}/{80}=0,034375(mol)`

`->m_{Cu}=0,034375.64=2,2(g)`

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

`->24x+27y=10-2,2=7,8(1)`

`n_{H_2}=x+1,5y={8,96}/{22,4}=0,4(2)`

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

`m_{Mg}=0,1.24=2,4(g)`

`m_{Al}=0,2.27=5,4(g)`