\(n_{NaOH}=\dfrac{50.20}{100.40}=0,25\left(mol\right)\)
\(n_{CuSO_4}=\dfrac{416.5}{100.160}=0,13\left(mol\right)\)
PTHH : CuSO4 + 2NaOH ---> Cu(OH)2 + Na2SO4
Ta thấy : \(\dfrac{0,25}{2}< 0,13\) => Spu NaOH hết; CuSO4 dư
Theo pthh : nCu(OH)2 = \(\dfrac{1}{2}n_{NaOH}=0,125\left(mol\right)\)
=> mCu(OH)2 = 98.0,125 = 12,25 (g)