\(n_{NaOH}=\frac{200.10}{100.40}=0,5\left(mol\right)\\ PTHH:2NaOH+CuSO_4\rightarrow Cu\left(OH\right)_2+Na_2SO_4\\ m_{ddCuSO_4}=\frac{\left(0,25.160\right).100}{16}250\left(g\right)\\ m_{kt}=98.0,25=24,5\left(g\right)\\ C\%_{ddNa_2SO_4}=\frac{0,25.142}{200+250}.100\%=7,89\left(\%\right)\)