\(n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\\ \text{Đ}\text{ặt}:n_{CaO}=a\left(mol\right);n_{ZnO}=b\left(mol\right)\left(a,b>0\right)\\ a,CaO+2HCl\rightarrow CaCl_2+H_2O\\ ZnO+2HCl\rightarrow ZnCl_2+H_2O\\ \Rightarrow\left\{{}\begin{matrix}56a+81b=12,1\\2a+2b=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,002\left(mol\right)\\b=0,148\left(mol\right)\end{matrix}\right.\\ \Rightarrow\%m_{CaO}=\dfrac{0,002.56}{12,1}.100\approx0,926\%\\ \Rightarrow\%m_{ZnO}\approx99,074\%\)
b) Sao lại HCl thu được nhỉ?
\(n_{HCl}=\dfrac{10,95}{36,5}=0,3\\ (n_{CaO};n_{ZnO})=(x;y)\\ \Rightarrow 56x+81y=12,1(1)\\ CaO+2HCl\to CaCl_2+H_2O\\ ZnO+2HCl\to ZnCl_2+H_2O\\ \Rightarrow 2x+2y=0,3(2)\\ (1)(2)\Rightarrow \begin{cases} x=0,002(mol)\\ y=0,148(mol) \end{cases}\Rightarrow \begin{cases} \%_{CaO}=\dfrac{0,002.56}{12,1}.100\%=0,93\%\\ \%_{ZnO}=100\%-0,93\%=99,07\% \end{cases}\)
Câu b ko hiểu đề :v
