PTHH: \(CuO+2HCl\rightarrow CuCl_2+H_2O\) (1)
\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\) (2)
a) Ta có: \(n_{HCl}=0,1\cdot1,5=0,15\left(mol\right)\)
Gọi số mol của CuO là \(a\) \(\Rightarrow n_{HCl\left(1\right)}=2a\)
Gọi số mol của ZnO là \(b\) \(\Rightarrow n_{HCl\left(2\right)}=2b\)
Ta có hệ phương trình:
\(\left\{{}\begin{matrix}2a+2b=0,15\\80a+81b=6,05\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,025\\b=0,05\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{CuO}=0,025mol\\n_{ZnO}=0,05mol\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{CuO}=0,025\cdot80=2\left(g\right)\\m_{ZnO}=4,05\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{CuO}=\frac{2}{6,05}\cdot100\approx33,06\%\\\%m_{ZnO}\approx66,94\%\end{matrix}\right.\)
b) PTHH: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\) (3)
\(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\) (4)
Theo PTHH: \(\left\{{}\begin{matrix}n_{H_2SO_4\left(3\right)}=n_{CuO}=0,025mol\\n_{H_2SO_4\left(4\right)}=n_{ZnO}=0,05mol\end{matrix}\right.\)
\(\Rightarrow n_{H_2SO_4}=0,075mol\) \(\Rightarrow m_{H_2SO_4}=0,075\cdot98=7,35\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\frac{7,35}{20\%}=36,75\left(g\right)\)