a, PT: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
Ta có: \(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{CuO}=0,4\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,2}=2\left(M\right)\)
b, \(n_{CuCl_2}=n_{CuO}=0,2\left(mol\right)\Rightarrow m_{CuCl_2}=0,2.135=27\left(g\right)\)