$n_{HCl}=0,1.3=0,3mol$
$PTHH :$
$CaO+2HCl\to CaCl_2+H_2$
$CaCO3+2HCl\to CaCl_2+CO_2+H_2O$
Gọi $n_{CaO}=a;n_{CaCO_3}=b (a,b\in N*)$
Ta có :
$m_{hh}=72a+100b=11,7g$
$n_{HCl}=2a+2b=0,3mol$
Ta có hpt :
$\left\{\begin{matrix}
72a+100b=11,7 & \\
2a+2b=0,3 &
\end{matrix}\right.$
$\Leftrightarrow \left\{\begin{matrix}
a=0,12 & \\
b=0,03 &
\end{matrix}\right.$
Theo pt :
$n_{CaCl_2(1)}=n_{CaO}=a=0,12mol$
$n_{CaCl_2(2)}=n_{CaCO_3}=b=0,03mol$
$\Rightarrow m_{CaCl_2}=(0,12+0,03).111=16,65g$
CaO+2HCl->CaCl2+H2O
CaCO3+2HCl->CaCl2+H2O+CO2
nHCl=0,3 mol
ta có :56x+100y=11,7
2x+2y=0,3 mol
=>x=y=0,075
=>mCaCl2=0,15.111=16,65g
