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

$n_{HCl}$ = 1,2.0,35 = 0,42 (mol)

Fe + 2HCl → Fe$Cl_{2}$ + $H_{2}$

0,2    0,42                                        (mol)

$\frac{0,2}{1}$ < $\frac{0,42}{2}$

--> $n_{HCl}$ dư = 0,42 - 0,2.2 = 0,02 (mol)

a) m HCL dư = 0,02 . 36,5 = 0,73(g)

b) $n_{H_{2}}$ = $n_{Fe}$ = 0,2 (mol) 

$V_{H_{2}}$ = 0,2 . 22,4 = 4,48 (l)

c)  $C_{M_{HCl dư}}$ = 0,02/0,35 = $\frac{2}{35}$ (M)

$C_{M_{FeCl}}$ = 0,2/0,35 = $\frac{4}{7}$ (M)

Em tham khảo nha :

\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{Fe}} = \dfrac{{11,2}}{{56}} = 0,2mol\\
{n_{HCl}} = 0,35 \times 1,2 = 0,42mol\\
\dfrac{{0,2}}{1} < \dfrac{{0,42}}{2} \Rightarrow HCl\text{ dư}\\
{n_{HC{l_d}}} = 0,42 - 0,2 \times 2 = 0,02mol\\
{m_{HC{l_d}}} = 0,02 \times 36,5 = 0,73g\\
b)\\
{n_{{H_2}}} = {n_{Fe}} = 0,2mol\\
{V_{{H_2}}} = 0,2 \times 22,4 = 4,48l\\
c)\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,2mol\\
{C_{{M_{FeC{l_2}}}}} = \dfrac{{0,2}}{{0,35}} = \dfrac{4}{7}M\\
{C_{{M_{HC{l_d}}}}} = \dfrac{{0,02}}{{0,35}} = \dfrac{2}{{35}}M
\end{array}\)