a, PT: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
b, Ta có: \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{CuCl_2}=n_{CuO}=0,1\left(mol\right)\\n_{HCl}=2n_{CuO}=0,2\left(mol\right)\end{matrix}\right.\)
\(m_{CuCl_2}=0,1.135=13,5\left(g\right)\)
c, \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{7,3}{7,3\%}=100\left(g\right)\)