a) \(n_R=\dfrac{3,68}{M_R}\left(mol\right)\)
PTHH: 2R + 2HCl --> 2RCl + H2
\(\dfrac{3,68}{M_R}\)-------->\(\dfrac{3,68}{M_R}\)->\(\dfrac{1,84}{M_R}\)
=> \(\dfrac{3,68}{M_R}\left(M_R+35,5\right)=9,36\)
=> MR = 23 (g/mol)
=> R là Natri (Na)
b) \(n_{H_2}=\dfrac{1,84}{23}=0,08\left(mol\right)\)
=> VH2 = 0,08.22,4 = 1,792 (l)