bài 7 thoy ạ, ngắn gọn

$\text{$n_{Fe}$ = 0,05 mol}$

$\text{PTHH: Fe + 2HCl → $FeCl_{2}$ + $H_{2}$ ↑ (1)}$

$\text{từ phương trình (1)⇒$n_{H_2}$ = 0,05 mol}$

$\text{⇒$V_{H_2}$ = 1,12 lít}$

$\text{từ phương trình (1)⇒$n_{HCl}$ = 0,1 mol}$

$\text{⇒$m_{HCl}$ = 3,65 gam}$

$\text{⇒$m_{dung dịch HCl}$ = 3,65 : 10% = 36,5 gam}$

`a)\ Fe+2HCl->FeCl_2+H_2`

`b)`

`n_{Fe}={2,8}/{56}=0,05(mol)`

Theo PT: `n_{H_2}=n_{Fe}=0,05(mol)`

`->V_{H_2}=0,05.22,4=1,12(l)`

`c)`

Theo PT: `n_{HCl}=2n_{Fe}=0,1(mol)`

`->m_{HCl}=0,1.36,5=3,65(g)`

`->m_{dd\ HCl}={3,65}/{10\%}=36,5(g)`