$n_{Fe}=16,8/56=0,3mol$

$PTHH :$

$Fe+2HCl\to FeCl_2+H_2↑$

a/Theo pt :

$n_{HCl}=2.n_{Fe}=2.0,3=0,6mol$

$⇒m_{HCl}=0,6.26,5=21,9g$

$⇒C\%_{HCl}=\dfrac{m_{ct}}{m_{dd}}.100\%=\dfrac{21,9}{200}.100\%=10,95\%$

b/Theo pt :

$n_{H_2}=n_{Fe}=0,3mol$

$⇒V_{H_2}=0,3.22,4=6,72l$

$c/n_{FeCl_2}=n_{Fe}=0,3mol$

$⇒m_{FeCl_2}=0,3.127=38,1g$

$m_{H_2}=0,3.2=0,6g$

$m_{dd spư}=16,8+200-0,6=216,2g$

$⇒C\%_{FeCl_2}=\dfrac{m_{ct}}{m_{dd}}.100\%=\dfrac{38,1}{216,2}.100\%=17,62\%$

Đáp án:

 a) 10,95%

b) 6,72l

c) 17,62%

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
nFe = \dfrac{{16,8}}{{56}} = 0,3\,mol\\
nHCl = 2nFe = 0,6\,mol\\
C\% HCl = \dfrac{{0,6 \times 36,5}}{{200}} \times 100\%  = 10,95\% \\
b)\\
n{H_2} = nFe = 0,3\,mol \Rightarrow V{H_2} = 0,3 \times 22,4 = 6,72l\\
c)\\
m{\rm{dd}}spu = 16,8 + 200 - 0,3 \times 2 = 216,2g\\
C\% FeC{l_2} = \dfrac{{0,3 \times 127}}{{216,2}} \times 100\%  = 17,62\% 
\end{array}\)