a/ 2NaOH + CuSO4 -----> Na2SO4 + Cu(OH)2
b/ \(m_{CuSO_4}=200.b\%\) \(\Rightarrow n_{CuSO_4}=\frac{200.b\%}{160}=\frac{5.b\%}{4}\) (mol)
\(m_{NaOH}=4\%.150=6\left(g\right)\) \(\Rightarrow n_{NaOH}=\frac{6}{40}=0,15\left(mol\right)\)
Theo đề bài thì 2nNaOH = nCuSO4
\(\Rightarrow\frac{5}{4}.b\%=0,3\Rightarrow b\%=0,24\%\)