Bài 1: 

 a) PTHH: $Fe+2HCl→FeCl_2+H_2$

b) $n_{H_2}=\dfrac{22,4}{22,4}=1(mol)$

  $n_{Fe}=n_{H_2}=1(mol)$

⇒ $m_{Fe}=56×1=56(g)$

c) $n_{HCl}=2n_{Fe}=2×1=2(mol)$

  $m_{HCl}=2×36,5=73(g)$

⇒ $C$%$_{ddHCl}=\dfrac{73}{300}×100$% $=$ $24,3$%

 Bài 2:

a) PTHH: $Mg+2HCl→MgCl_2+H_2$

b) $n_{Mg}=\dfrac{2,4}{24}=0,1(mol)$

 $n_{HCl}=2n_{Mg}=2×0,1=0,2(mol)$

⇒ $m_{HCl}=0,2×36,5=7,3(g)$

c) $C$%$_{ddHCl}=\dfrac{7,3}{400}×100$% $=$ $1,825$%

Bạn tham khảo nha!

Bài 1: a. `-` `Fe + 2HCl -> FeCl_2 + H_2 ↑`

b. `-` $n_{H_2}$ `=` $\dfrac{22,4}{22,4}$ `= 1` `(mol)` 

`-` Theo phương trình $n_{Fe}$ `=` $n_{H_2}$ `= 1` `(mol)` 

`->` $m_{Fe}$ `= 1 xx 56 = 56` `(g)`

c. `-` Theo phương trình $n_{HCl}$ `= 2` `(mol)`

`->` $m_{HCl}$ `= 2 xx 36,5 = 73` `(g)`

`->` `C%_{HCl} =` $\dfrac{73}{300}$ `xx 100 = 24,33%` 

Bài 2: a. `-` `Mg + 2HCl -> MgCl_2 + H_2 ↑`

b. `-` $n_{Mg}$ `=` $\dfrac{2,4}{24}$ `= 0,1` `(mol)` 

`-` Theo phương trình $n_{HCl}$ `= 0,2` `(mol)` 

`->` $m_{HCl}$ `= 0,2 xx 36,5 = 7,3` `(g)`

c. `-` `C%_{HCl} =` $\dfrac{7,3}{400}$ `xx 100% = 1,825%`