a, $Fe+2HCl\to FeCl_2+H_2$

b, 

$n_{Fe}=\dfrac{11,2}{56}=0,2 mol$

$\Rightarrow n_{HCl}=2n_{Fe}=0,2.2=0,4 mol$

$m_{HCl}=0,4.36,5=14,6g$

c,

$m_{dd HCl}=14,6:7,3\%=200g$

d,

$n_{H_2}= n_{FeCl_2}=0,2 mol$ 

$m_{dd \text{spu}}= m_{Fe}+200-m_{H_2}= 11,2+200-0,2.2=210,8g$

$\Rightarrow C\%_{FeCl_2}=\dfrac{0,2.127.100}{210,8}=12,05\%$

Đáp án:

Giải thích các bước giải:

$a/$

$Fe + 2HCl → FeCl_2 + H_2$
$b/$

có : $n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)$

Theo PTHH , ta có :

$n_{HCl} = 2n_{Fe} = 0,2.2 = 0,4(mol)$
$⇒ m_{HCl} = 0,4.36,5 = 14,6(gam)$

$c/$

$m_{\text{dung dịch HCl}} = \dfrac{m_{HCl}}{C\%} = \dfrac{14,6}{7,3\%} =200(gam)$

Theo PTHH: $n_{H_2}=n_{FeCl_2} = n_{Fe} = 0,2(mol)$

Sau phản ứng :

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

$= 11,2 + 200 - 0,2.2 = 210,8(gam)$

$⇒ C\%_{FeCl_2} = \dfrac{0,2.127}{210,8}.100\% = 12,05\%$