Cu k phản ứng với HCl
\(\rightarrow m_{Cu}=1\left(g\right)\)
\(PTHH:CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(n_{CuO}=\frac{5-1}{80}=0,05\left(mol\right)\)
\(n_{HCl}=1\left(mol\right)\)
\(TL:\frac{1}{2}>\frac{0,05}{1}\rightarrow HCl.du\)
\(C_{M_A}=\frac{0,05}{0,5}=0,1\left(M\right)\)
\(C_{M_{HCl.du}}=\frac{1-0,05.2}{0,5}=1,8\left(M\right)\)