\(\begin{array}{l}
1)\\
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
nMg = \dfrac{{4,8}}{{24}} = 0,2\,mol\\
 \Rightarrow n{H_2} = nMg = 0,2\,mol \Rightarrow V{H_2} = 0,2 \times 22,4 = 4,48l\\
b)\\
nHCl = 2nMg = 0,4\,mol\\
{C_M}HCl = \dfrac{{0,4}}{{0,2}} = 2M\\
2)\\
mNaCl = 200 \times 3\%  = 6g\\
m{H_2}O = 200 - 6 = 194g
\end{array}\)

Cho 6 g NaCl vào 194 g nước ta thu được dd NaCl 3%

\(\begin{array}{l}
3)\\
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
b)\\
nFe = \dfrac{{22,4}}{{56}} = 0,4\,mol\\
n{H_2} = nFe = 0,4\,mol \Rightarrow V{H_2} = 0,4 \times 22,4 = 8,96l\\
c)\\
m{\rm{dd}}spu = 22,4 + 200 - 0,4 \times 2 = 221,6g\\
C\% FeC{l_2} = \dfrac{{0,4 \times 127}}{{221,6}} \times 100\%  = 22,92\% 
\end{array}\)