\(\begin{array}{l} PTHH:\\ CaO+2HCl\to CaCl_2+H_2O\ (1)\\ CaCO_3+2HCl\to CaCl_2+CO_2↑+H_2O\ (2)\\ n_{HCl}=0,1\times 3=0,3\ mol.\\ \text{Gọi $n_{CaO}$ là a (mol), $n_{CaCO_3}$ là b (mol).}\\ \text{Theo đề bài ta có hệ pt:}\ \begin{cases}56a+100b=11,7\\2a+2b=0,3\end{cases}\ \Rightarrow \begin{cases}a=0,075\\b=0,075\end{cases}\\ \Rightarrow \sum n_{CaCl_2}=n_{CaO}+n_{CaCO_3}=0,075+0,075=0,15\ mol.\\ \Rightarrow m_{CaCl_2}=0,15\times 111=16,65\ g.\end{array}\)
