$n_{CO_2}=\dfrac{0,672}{22,4}=0,03(mol)$
Gọi chung 2 muối cacbonat là $R_2(CO_3)_n$
$R_2(CO_3)_n+ 2nHCl\to 2RCl_n+ nCO_2+nH_2O$
Theo PTHH:
$n_{HCl}=2n_{CO_2}=0,06(mol)$
$n_{H_2O}=n_{CO_2}=0,03(mol)$
BTKL:
$m_{\text{muối}}=10+0,06.36,5-0,03.44-0,03.18=10,33g$