\(PTHH:CuO+2HCl\rightarrow CuCl_2+H_2O\)
+ \(\dfrac{n_{CuO\left(bđ\right)}}{n_{CuO\left(PTHH\right)}}=\dfrac{0,25}{1}=0,25\)
+ \(\dfrac{n_{HCl\left(bđ\right)}}{n_{HCl\left(PTHH\right)}}=\dfrac{0,5}{2}=0,25\)
=> pư trên vừa đủ
=> \(m_{CuCl_2}=0,25.135=33,75g\)