PTHH: CuO + 2HCl --> CuCl2 + H2O
\(n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right)\)
\(n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,25}{1}=\dfrac{0,5}{2}\) => pư vừa đủ
PTHH: CuO + 2HCl --> CuCl2 + H2O
0,25-->0,5------>0,25
=> \(m_{CuCl_2}=0,25.135=33,75\left(g\right)\)
