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

PTHH: $Fe + 2HCl \rightarrow FeCl_2 + H_2$

            0,2--->0,4-------------->0,2------>0,2

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

b, $m_{ddHCl} = \dfrac{0,4.36,5}{10\%} = 146 (g)$

=> $m_{dd} = 146 - 0,2.2 + 11,2 = 156,8 (g)$

=> $C\% = \dfrac{0,2.127}{156,8} . 100\% = 16,07\%$

Em tham khảo nha :

\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{Fe}} = \dfrac{{11,2}}{{56}} = 0,2mol\\
{n_{{H_2}}} = {n_{Fe}} = 0,2mol\\
{V_{{H_2}}} = 0,2 \times 22,4 = 4,48l\\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,4mol\\
{m_{HCl}} = 0,4 \times 36,5 = 14,6g\\
{m_{{\rm{dd}}HCl}} = \dfrac{{14,6 \times 100}}{{10}} = 146g\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,2mol\\
{m_{FeC{l_2}}} = 0,2 \times 127 = 25,4g\\
C{\% _{FeC{l_2}}} = \dfrac{{25,4}}{{11,2 + 146 - 0,2 \times 2}} \times 100\%  = 16,2\% 
\end{array}\)