\(n_{H_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)
\(\overline{M}+2HCl\rightarrow\overline{M}Cl_2+H_2\)
\(0.15..............................0.15\)
\(M_{\overline{M}}=\dfrac{2.25}{0.15}=15\left(\dfrac{g}{mol}\right)\)
\(A< 15< B\)\(\)
\(A=9\Rightarrow A:Be\)
\(B=24\Rightarrow B:Mg\)
\(n_{HCl}=2n_{H_2}=2\cdot0.15=0.3\left(mol\right)\)
\(BTKL:\)
\(m_{KL}+m_{HCl}=m_M+m_{H_2}\)
\(\Rightarrow m_M=2.25+0.3\cdot36.5-0.15\cdot2=12.9\left(g\right)\)