a,

Gọi $x$, $y$ là số mol $CaCO_3$, $MgCO_3$

$CaCO_3\buildrel{{t^o}}\over\to CaO+CO_2$

$MgCO_3\buildrel{{t^o}}\over\to MgO+CO_2$

$\Rightarrow x+y=0,2$               (1)

Ta có $m_{CaCO_3}=0,8m (g)$, $m_{MgCO_3}=0,084m (g)$

$\Rightarrow \dfrac{m_{CaCO_3}}{m_{MgCO_3}}=\dfrac{0,8}{0,084}=\dfrac{200}{21}$

$\Rightarrow \dfrac{100x}{84y}=\dfrac{200}{21}$

$\Leftrightarrow 100.21x-200.84y=0$  (2)

$(1)(2)\Rightarrow x=\dfrac{8}{45}(mol); y=\dfrac{1}{45}(mol)$

$\Rightarrow m=100x:80\%=\dfrac{200}{9}(g)$

b,

Chất rắn sau phản ứng gồm:

$CaO$: $\dfrac{8}{45}$ mol

$MgO$: $\dfrac{1}{45}$ mol

$Al_2O_3$: $\dfrac{10,2\%m}{102}=\dfrac{1}{45}$ mol

Tạp chất $m.1,4\%=\dfrac{11}{45}g$

$\Rightarrow m_Y=\dfrac{11}{45}(g)$

$CaO+2HCl\to CaCl_2+H_2O$

$MgO+2HCl\to MgCl_2+H_2O$

$Al_2O_3+6HCl\to 2AlCl_3+3H_2O$

$\Rightarrow n_{CaCl_2}=\dfrac{8}{45}(mol)$; $n_{MgCl_2}=\dfrac{1}{45}(mol)$; $n_{AlCl_3}=\dfrac{2}{45}(mol)$

$\to a=111.\dfrac{8}{45}+95.\dfrac{1}{45}+133,5.\dfrac{2}{45}=27,77(g)$