Gọi x, y là số mol $CaCO_3$, $MgCO_3$.
$n_{CO_2}=\dfrac{2,296}{22,4}=0,1025(mol)$
$CaCO_3+2HCl\to CaCl_2+CO_2+H_2O$
$MgCO_3+2HCl\to MgCl_2+CO_2+H_2O$
$\Rightarrow x+y=0,1025$ (1)
$MgCl_2+2NaOH\to Mg(OH)_2+2NaCl$
$Mg(OH)_2\buildrel{{t^o}}\over\to MgO+H_2O$
$\Rightarrow n_{MgO}=y (mol)$
$\Rightarrow 40y=2,4$ (2)
$(1)(2)\Rightarrow x=0,0425; y=0,06$
$\%m_{CaCO_3}=\dfrac{0,0425.100.100}{10}=42,5\%$
$\%m_{MgCO_3}=\dfrac{0,06.84.100}{10}=50,4\%$