a,
\(n_{HCl}=\frac{200.20\%}{36,5}=1,1\left(mol\right)\)
\(A+2HCl\rightarrow ACl_2+H_2\)
\(\Rightarrow n_A=n_{H2}=0,2\left(mol\right)\)
\(\Rightarrow M_A=\frac{4,8}{0,2}=24\)
Vậy A là magie
b,
Sau phản ứng tạo 0,2 mol MgCl2; dư 0,7 mol HCl
\(m_{dd_{spu}}=4,8+200-0,2.2=204,4\left(g\right)\)
\(\Rightarrow C\%_{MgCl2}=\frac{0,2.95.100}{204,4}=9,3\%\)
\(\Rightarrow C\%_{HCl}=\frac{0,7.36,5.100}{204,4}=12,5\%\)