$M_2(CO_3)_n+2nHCl\to 2MCl_n+nCO_2+nH_2O$
Xét 1 mol muối cacbonat.
$\Rightarrow n_{HCl}=2n$, $n_{MCl_n}=2$, $n_{CO_2}=n$ $(mol)$
$m_{dd HCl}=2n.36,5:7,3\%=1000n(g)$
$\Rightarrow m_{dd \text{spứ}}=(2M+60n).1+1000n-44n=2M+1016n(g)$
$m_{MCl_n}=2(M+35,5n)=2M+71n(g)$
$\Rightarrow 2M+71n=0,11017(2M+1016n)$
$\Leftrightarrow M=23n$
$n=1\Rightarrow M=23(Na)$
Vậy kim loại là natri.