-$n_{hh...khí}=\frac{5,6}{22,4}=0,25(mol)$
$M_2CO_3+2HCl→2MCl+CO_2↑+H_2O$
$M_2SO_3+2HCl→2MCl+SO_2↑+H_2O$
-Vì $n_{CO_2}=n_{hh...muối}=0,25(mol)$
$⇒M_{hh...muối}=\frac{29,5}{0,25}=118(g/mol)$
$⇒M_{M_2CO_3}<M_{hh...muối}<M_{M_2SO_3}$
$⇔2.M+12+16.3<118<2.M+32+16.3$
$⇔2.M+60<118<2.M+80$
$⇔19<M<29$
⇒ $M =23 (g/mol)$
⇒ $M$ là nguyên tố Natri($Na$)
