\(X+2H_2O\rightarrow X\left(OH\right)_2+H_2\)
\(m_{dd_{H2O}}=73.1=73\left(g\right)\)
\(n_X=\frac{27,4}{M_X}\left(mol\right)\)
\(n_{H2}=n_{X\left(OH\right)2}=n_X=\frac{27,4}{M_X}\left(mol\right)\)
\(m_{H2}=\left(\frac{27,4}{M_X}\right).2=\frac{54,8}{M_X}\left(g\right)\)
\(\Rightarrow m_{dd_{spu}}=73+27,4-\frac{54,8}{M_X}=100,4-\frac{54,8}{M_X}\left(g\right)\)
\(m_{X\left(OH\right)2}=\left(\frac{27,4}{M_X}\right).X+34=\frac{\left(27,4X+931,6\right)}{X}\left(g\right)\)
\(C\%_{X\left(OH\right)2}=\frac{27,4X+931,9}{X}:\left(100-\frac{54,8}{M_X}\right).100\%=34,2\)
\(\Rightarrow X=137\left(\frac{g}{mol}\right)\)
Vậy X là Bari (Ba)