$n_{NO\ ban đầu} = \dfrac{21,96}{22,4} : 98\% = 1(mol)$
$3M +N_2 \xrightarrow{t^o} M_3N_2$
$M_3N_2 + 6H_2O \to 3M(OH)_2 + 2NH_3$
$2NH_3 + \dfrac{5}{2}O_2 \xrightarrow{t^o,Pt} 2NO + 3H_2O$
$n_{NH_3} = n_{NO} = 1(mol)$
$n_{M_3N_2} = \dfrac{1}{2}n_{NH_3} = 0,5(mol)$
$n_M = 3n_{M_3N)2} = 0,5.3 = 1,5(mol)$
$\Rightarrow M = \dfrac{60}{1,5} = 40(Ca)$
Do đó, M là Canxi