a) \(n_{H_2SO_4}=0,2.2=0,4\left(mol\right)\)
\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
Đặt \(n_{Fe_2O_3}=a\left(mol\right);n_{CuO}=b\left(mol\right)\)
Ta có:
\(\left\{{}\begin{matrix}160a+80b=24\\3a+b=0,4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)
\(\%m_{Fe_2O_3}=\dfrac{160.0,1}{24}.100\%=66,67\%\\ \%m_{CuO}=100\%-66,67\%=33,33\%\)
b) \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(n_{HCl}=6.0,1+2.0,1=0,9\left(mol\right)\)
\(m_{HCl}=0,9.36,5=32,85\left(g\right)\)
\(m_{ddHCl}=\dfrac{32,85.100}{14,7}=223,47\left(g\right)\)
