Đáp án:

 7,3% và 12,05%

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
b)\\
nFe = \dfrac{m}{M} = \dfrac{{2,8}}{{56}} = 0,05\,mol\\
nHCl = 2nFe = 0,05 \times 2 = 0,1\,mol\\
mHCl = n \times M = 0,1 \times 36,5 = 3,65g\\
C\% HCl = \dfrac{{mct}}{{m{\rm{dd}}}} \times 100\%  = \dfrac{{3,65}}{{50}} \times 100\%  = 7,3\% \\
c)\\
n{H_2} = nFe = 0,05\,mol\\
m{\rm{dd}}spu = 2,8 + 50 - 0,05 \times 2 = 52,7g\\
C\% FeC{l_2} = \dfrac{{mct}}{{m{\rm{dd}}}} \times 100\%  = \dfrac{{0,05 \times 127}}{{52,7}} \times 100\%  = 12,05\% 
\end{array}\)

$n_{Fe}=2,8/56=0,05mol$

$a/Fe+2HCl\to FeCl_2+H_2↑$

$\text{b/Theo pt :}$

$n_{HCl}=2.n_{Fe}=2.0,05=0,1mol$

$⇒m_{HCl}=0,1.36,5=3,65g$

$⇒C\%_{HCl}=\dfrac{3,65}{50}.100\%=7,3\%$

$\text{c/Theo pt :}$

$n_{FeCl_2}=n_{Fe}=0,05mol$

$⇒m_{FeCl_2}=0,05.127=6,35g$

$m_{ddspu}=2,8+50-0,05.2=52,7g$

$⇒C\%_{FeCl_2}=\dfrac{6,35}{52,7}.100\%=12,05\%$