Câu IV:
\(1,PTHH:Na_2O+H_2O\rightarrow2NaOH\\ 2NaOH+CuSO_4\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\\ m_{NaOH}=\dfrac{100\cdot8\%}{100\%}=8\left(g\right)\\ \Rightarrow n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\\ \Rightarrow n_{Cu\left(OH\right)_2}=\dfrac{1}{2}n_{NaOH}=0,1\left(mol\right)\\ \Rightarrow m_{Cu\left(OH\right)_2}=0,1\cdot98=9,8\left(g\right)\\ 2,PTHH:Cu\left(OH\right)_2+2HCl\rightarrow CuCl_2+H_2O\\ \Rightarrow n_{HCl}=2n_{Cu\left(OH\right)_2}=0,2\left(mol\right)\\ \Rightarrow m_{CT_{HCl}}=0,2\cdot36,5=7,3\left(g\right)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{7,3\cdot100\%}{25\%}=29,2\left(g\right)\)
