$n_{Fe}$= $\frac{m_{Fe}}{M_{Fe}}$= $\frac{11,2}{56}$= 0,2 (mol)

$Fe_{}$ + 2$HCl_{}$ → $FeCl_{2}$ + $H_{2}$

a) Theo phương trình ta có$n_{H_2}$ = $n_{Fe}$= $n_{FeCl_2}$ =0,2 mol

$n_{HCl}$ = 2. $n_{Fe}$=2. 0,2=0,4 mol

a)$V_{H_2}$ = $n_{H_2}$. 22,4= 0,2. 22,4= 4,48(L)

b)$m_{HCl}$ = $n_{HCl}$ . $M_{HCl}$ = 0,4 . 36,5= 14,6(g)

C% = $\frac{m_{HCl}}{m{dung dịch HCl}}$. 100%= $\frac{14,6}{200}$ .100%= 7,3%

c) Ta có: $m_{dung dịch SP Ư}$= $m_{Fe}$ + $m_{dung dịch HCl}$ - $m_{H_2}$ 

⇒ $m_{dung dịch SP Ư}$= 11,2+200-(0,2.2)=210,8(g)

$m_{FeCl_2}$= $n_{FeCl_2}$. $M_{FeCl_2}$= 0,2.127=25,4(g)

C%= $\frac{m_{FeCl_2}}{m_{dung dịch SP Ư}}$.100%= $\frac{25,4}{210,8}$.100% = 12,05%

Chúc bạn học tốt :))

`n_{Fe}=\frac{11,2}{56}=0,2(mol)`

`a)` `Fe+2HCl\to FeCl_2+H_2`

`n_{Fe}=n_{H_2}=0,2(mol)`

`=> V_{H_2}=0,2.22,4=4,48(l)`

`b)` `n_{HCl}=2n_{H_2}=0,4(mol)`

`=>C%_{HCl}=\frac{0,4.36,5.100%}{200}=7,3%`

`c)` BTKL:

`m_{\text{dd spu}}=m_{Fe}+m_{\text{dd HCl}}-m_{H_2}`

`=> m_{\text{dd spu}}=11,2+200-0,2.2=210,8g`

`C%_{FeCl_2}=\frac{127.0,2.100%}{210,8}\approx 12,05%`