a) \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
b) \(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)
Theo PTHH: \(n_{CuCl_2}=n_{CuO}=0,05\left(mol\right)\Rightarrow m_{CuCl_2}=0,05.135=6,75\left(g\right)\)
c) mdd sau pư = 156 + 4 = 160 (g)
\(C\%_{CuCl_2}=\dfrac{6,75}{160}.100\%=4,21875\%\)