a)\(CuO+H2SO4-->CuSO4+H2O\)
x-----------x(mol)
\(FeO+H2SO4-->FeSO4+H2O\)
y------y(mol)
\(n_{H2SO4}=0,2.1=0,2\left(mol\right)\)
Theo bài ta có hpt
\(\left\{{}\begin{matrix}80x+72y=15\\x+y=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,075\\y=0,125\end{matrix}\right.\)
\(\%m_{CuO}=\frac{0,075.80}{15}.100\%=40\%\)
\(\%m_{FeO}=100-40=60\%\)
b)\(CuO+2HCl-->CuCl2+H2O\)
0,075------0,15(mol)
\(FeO+2HCl--.FeCl2+H2O\)
0,125--->0,25(mol)
\(\sum n_{HC_{ }l}=0,25+0,15=0,4\left(mol\right)\)
\(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(m_{ddHCl}=\frac{14,6.100}{30}=48,667\left(g\right)\)
Đề thiếu CM của H2SO4 nha bạn