$n_{CO_2}=\dfrac{2,24}{22,4}=0,1(mol)$
Gọi CTHH muối cacbonat là $R_2(CO_3)_n$
$R_2(CO_3)_n+2nHCl\to 2RCl_n+nCO_2+nH_2O$
$\Rightarrow n_{R_2(CO_3)_n}=\dfrac{0,1}{n}$
$M=\dfrac{10n}{0,1}=100n=2R+60n$
$\Leftrightarrow R=20n$
$n=2\Rightarrow R=40(Ca)$
Vậy muối cacbonat là $CaCO_3$
$CaCO_3+2HCl\to CaCl_2+CO_2+H_2O$
$n_{HCl}=0,1.2=0,2(mol)$
$\Rightarrow V_{HCl}=\dfrac{0,2.36,5:20\%}{1,1}=33,2ml$