Question

Answer 1

\(n_{H_2}=\dfrac{3,6}{24}=0,15\left(mol\right)\)

PTHH: 2M + 2xH₂O ---> 2M(OH)_x + xH₂

\(n_M=\dfrac{2}{x}n_{H_2}=\dfrac{2}{x}.0,15=\dfrac{0,3}{x}\left(mol\right)\\ \rightarrow M_M=\dfrac{20,55}{\dfrac{0,3}{x}}=68,5x\left(\dfrac{g}{mol}\right)\)

x = 2 thoả mãn => M_M = 137 (g/mol) => M là Ba

Answer 2

nH2 = 3,6 :24 =0,15 (mol)
pthh : M + yH2O -> H2 + M(OH)y 
          0,15<--------0,15 (mol) 
=> M_M = 20,55 : 0,15 = 137 (g/mol) 
  => M là Bari