a) \(A+2HCl\xrightarrow[]{}ACl_2+H_2\)
\(n_{ACl_2}=n_A=\dfrac{3,6}{A}\left(mol\right)\)
\(M_A=\dfrac{14,2}{\dfrac{3,6}{A}}=3,94A\)
hay \(A+35,5.2=3,94A\)
⇒ \(A=24\)
Vậy A là Mg
b) \(Mg+2HCl\xrightarrow[]{}MgCl_2+H_2\)
\(n_{HCl}=2n_{Mg}=2.\dfrac{3,6}{24}=0,3\left(mol\right)\)
\(C\%_{HCl}=\dfrac{0,3.36,5}{200}.100=5,475\%\)