`n_{Fe}=(16,8)/(56)=0,3(mol)`

`a,`

PTHH:

`Fe+2HCl->FeCl_2+H_2`

`0,3->0,6->0,3->0,3(mol)`

`b,`

Từ PTHH : `n_{H_2}=0,3(mol)`

`=>V_{H_2}=0,3.22,4=6,72(lít)`

`c,`

`n_{CuO}=(20)/(80)=0,25(mol)`

PTHH:

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

Lập tỉ lệ :

`(n_{CuO})/1 = (0,25)/1 < (n_{H_2})/1 = (0,3)/1`

`=>` Sau phản ứng `CuO` hết, `H_2` dư

`=>n_{Cu}=n_{CuO}=0,25(mol)`

`=>m_{Cu}=0,25.64=16(g)`

$a,$ 

$n_{Fe}=\dfrac{16,8}{56}=0,3(mol)$

 $Fe+2HCl→FeCl_2+H_2$

$b,$

 Theo PT: $n_{H_2}=n_{Fe}=0,3(mol)$

$→V_{H_2}=0,3×22,4=6,72(l)$

$c,$

$n_{CuO}=\dfrac{20}{80}=0,25(mol)$

 $CuO+H_2\xrightarrow{t^o}Cu+H_2O$

Xét $\dfrac{0,25}{1}<\dfrac{0,3}{1}$ $⇒\ H_2$ dư

Theo PT: $n_{Cu}=n_{CuO}=0,25(mol)$

$→m_{Cu}=0,25×64=16(g)$