\(4.CaO+H_2O\rightarrow Ca\left(OH\right)_2\\ Ca\left(OH\right)_2+Na_2CO_3\rightarrow CaCO_3+2NaOH\\ 2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ Na_2SO_4+BaCl_2\rightarrow BaSO_4+2NaCl\)
\(5.a.n_{BaCl_2}=\dfrac{400.5,2\%}{208}=0,1\left(mol\right)\\ n_{CuSO_4}=\dfrac{200.12\%}{160}=0,15\left(mol\right)\\CuSO_4+BaCl_2\rightarrow CuCl_2+BaSO_4 \\ LTL:\dfrac{0,15}{1}>\dfrac{0,1}{1}\Rightarrow CuSO_4dư\\ m_{BaSO_4}=n_{BaCl_2}=0,1\left(mol\right)\\ \Rightarrow m_{BaSO_{\text{4 }}}=23,3\left(g\right)\\ b.m_{ddsaupu}=400+200-23,3=576,7\left(g\right)\\ n_{CuSO_4\left(Dư\right)}=0,15-0,1=0,05\left(mol\right)\\ n_{CuCl_2}=n_{BaCl_2}=0,1\left(mol\right)\\ C\%_{CuSO_4}=\dfrac{0,05.160}{576,7}.100=1,39\%\\ C\%_{CuCl_2}=\dfrac{0,1.135}{576,7}.100=2,34\%\)
