a,
$n_{CO_2}=\dfrac{2,24}{22,4}=0,1(mol)$
$CaCO_3+2HCl\to CaCl_2+CO_2+H_2O$
$\to n_{HCl\text{dư}}=2n_{CO_2}=0,2(mol)$
$n_{CaCl_2}=n_{CO_2}=0,1(mol)$
Đặt công thức oxit là $RO$
$RO+2HCl\to RCl_2+H_2O$
$m_{RCl_2}=30,1-0,1.111=19g$
Theo PT, ta có $n_{RO}=n_{RCl_2}$
$\to \dfrac{8}{M_R+16}=\dfrac{19}{M_R+71}$
$\to M_R=24(Mg)$
Vậy oxit là $MgO$
b,
$n_{HCl\text{pứ}}=2n_{MgO}=2.\dfrac{8}{40}=0,4(mol)$
$\to \sum n_{HCl}=0,4+0,2=0,6(mol)$
$\to C_{M_{HCl}}=\dfrac{0,6}{0,3}=2M$