\(R+2HCl\rightarrow RCl_2+H_2\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ n_R=n_{H_2}=0,1\left(mol\right)\\ M_R=\dfrac{6,5}{0,1}=65\left(\dfrac{g}{mol}\right)\\ \Rightarrow R\left(II\right):Kẽm\left(Zn=65\right)\)
R+ 2HCl →RCl2 + H2
0,1 ← 0,1 mol
n H2 = 2,24:22,4=0,1 mol
n R =6,5/MR
=> 6,5/MR =0,1 => MR =65
=> R là kẽm (Zn)