$a,PTPƯ:Fe+2HCl\xrightarrow{} FeCl_2+H_2↑$

$n_{Fe}=\dfrac{5,6}{56}=0,1mol.$

$Theo$ $pt:$ $n_{H_2}=n_{Fe}=0,1mol.$

$⇒V_{H_2}=0,1.22,4=2,24l.$

$b,Theo$ $pt:$ $n_{HCl}=2n_{Fe}=0,2mol.$

$⇒m_{HCl}=0,2.36,5=7,3g.$

$⇒m_{ddHCl}=\dfrac{7,3}{20\%}=36,5g.$

$c,m_{H_2}=0,1.2=0,2g.$

$⇒m_{dd spư}=m_{Fe}+m_{ddHCl}-m_{H_2}$

$⇒m_{dd spư}=5,6+36,5-0,2=41,9g.$

$Theo$ $pt:$ $n_{FeCl_2}=n_{Fe}=0,1mol.$

$⇒m_{FeCl_2}=0,1.127=12,7g.$

$⇒C\%_{FeCl_2}=\dfrac{12,7}{41,9}.100\%=30,31\%$

chúc bạn học tốt!

\(\begin{array}{l}
a)\\
Fe+2HCl\to FeCl_2+H_2\\
n_{H_2}=n_{Fe}=\frac{5,6}{56}=0,1(mol)\\
V_{H_2}=0,1.22,4=2,24(l)\\
b)\\
n_{HCl}=2n_{H_2}=0,2(mol)\\
m_{dd\,HCl}=\frac{0,2.36,5}{20\%}=36,5(g)\\
c)\\
n_{FeCl_2}=n_{Fe}=0,1(mol)\\
C\%_{FeCl_2}=\frac{0,1.127}{5,6+36,5-0,1.2}.100\%\approx 30,31\%
\end{array}\)