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 $m.1,4\%=\dfrac{11}{45}g$
$\Rightarrow m_Y=\dfrac{11}{45}(g)$
$CaO+2HCl\to CaCl_2+H_2O$
$MgO+2HCl\to MgCl_2+H_2O$
$Al_2O_3+6HCl\to 2AlCl_3+3H_2O$
$\Rightarrow n_{CaCl_2}=\dfrac{8}{45}(mol)$; $n_{MgCl_2}=\dfrac{1}{45}(mol)$; $n_{AlCl_3}=\dfrac{2}{45}(mol)$
$\to a=111.\dfrac{8}{45}+95.\dfrac{1}{45}+133,5.\dfrac{2}{45}=27,77(g)$
