ZnO + 2HCl -> ZnCl2 + H2O
x............2x
CuO + 2HCl -> CuCl2 + H2O
y..............2y
Theo bài ra :\(\left\{{}\begin{matrix}81x+80y=12,1\\2x+2y=0,3\end{matrix}\right.\) => x= 0,1 ; y =0,05 (mol)
=> %mCuO = 66,12%
% mZnO = 33,88%
b , \(\Sigma n_{H2SO4}=0,15\left(mol\right)\) => mH2SO4 =14,7 (g)
=> mdd = 14,7/19,6%=75(g)
\(n_{CuO}=x;n_{ZnO}=y\\ PTHH:CuO+2HCl\rightarrow CuCl_2+H_2O\\ PTHH:ZnO+2HCl\rightarrow ZnCl_2+H_2O\\ hpt:\left\{{}\begin{matrix}80x+81y=12,1\\2x+2y=1.0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\%m_{CuO}=\frac{80.0,05}{12,1}.100\%=33,06\left(\%\right)\\\%m_{ZnO}=100-33,06=66,94\left(\%\right)\end{matrix}\right.\)
\(PTHH:CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ PTHH:ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\\ m_{H_2SO_4}=98.\left(0,05+0,1\right)=14,7\left(g\right)\\ \Rightarrow m_{dd}=\frac{14,7.100}{19,6}=75\left(g\right)\)
