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câu 6:

`n_{H_2}={3,719}/{22,4}~~0,17(mol)`

`PTHH:Fe+2HCl->FeCl_2+H_2`

`n_{HCl}=2n_{H_2}=0,17.2=0,34(mol)`

`m_{HCl}=0,34.36,5=12,41(g)`

câu 7:

`n_{Zn}={8,45}/65=0,13(mol)`

`n_{Cl_2}={6,1975}/{24,79}=0,25(mol)`

`PTHH:Zn+Cl_2 \overset(t^o)(->)ZnCl_2`

 ta có: `{0,13}/1 <{0,25}/1 ->Cl_2` dư

$PTHH: Fe +2HCl \xrightarrow {} FeCl_2 +H_2$

$n_{H_2}=\dfrac{3,719}{24,79} ≈0,15(mol)$

$n_{HCl}=2n_{H_2}=0,3(mol)$

$→m_{HCl}=0,3 ×36,5 =10,95(g)$

Câu 7

$PTHh : Zn +Cl_2 \xrightarrow{to} ZnCl_2$

$n_{Cl_2}=\dfrac{6,1975}{24,79}=0,25(mol)$

$n_{Zn}=\dfrac{8,45}{65}=0,13(mol)$

Tỉ lệ $\dfrac{0,13}{1}<\dfrac{0,25}{1} $⇒ Zn Hết 

Chất còn dư là $Cl_2$