nCuSO4=0,2(mol)
nNaOH=0,2(mol)
pthh: CuSO4+2NaOH\(\rightarrow\) Cu(OH)2\(\downarrow\) +Na2SO4
xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,2}{2}\Rightarrow CuSO\text{4}dư\)
-Dung dịc A:Na2SO4, CuSO4
-Chất rắn B: Cu(OH)2
Theo pthh: nNa2SO4=nCuSO4 pứ=1/2nNaOH=0,1(mol)
\(\Rightarrow\)....