\(n_{CuSO_4}=\dfrac{4}{160}=0,025mol\)
\(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\)
0,025 0,05 0,025 0,025
\(m_{Cu\left(OH\right)_2\downarrow}=0,025\cdot98=2,45\left(g\right)\)
\(m_{ctNaOH}=0,05\cdot40=2\left(g\right)\)
\(\Rightarrow m_{ddNaOH}=\dfrac{2}{40}\cdot100=5\left(g\right)\)