2NaOH+ CuSO4-->Na2SO4 +Cu(OH)2↓(1)
Cu(OH)2--->CuO+H2O(2)
Ta có
n\(_{NaOH}=\frac{20}{40}=0,5\left(mol\right)\)
Theo pthh1
n\(_{Cu\left(OH\right)2}=\frac{1}{2}n_{NaOH}=0,25\left(mol\right)\)
Theo pthh2
n\(_{CuO}=n_{Cu\left(OH\right)2}=0,25\left(mol\right)\)
m\(_{CuO}=0,25.80=20\left(g\right)\)
nNaOH = 20/40=0.5 mol
2NaOH + CuSO4 --> Na2SO4 + Cu(OH)2
0.5___________________________0.25
Cu(OH)2 -to-> CuO + H2O
0.25__________0.25
mCuO = 0.25*80=20 g
Ta có: \(nNaOH=0,5\left(mol\right)\)
\(\text{PTHH : CuSO4+ NaOH}\rightarrow Na2SO4+Cu\left(OH\right)2\)
Theo PT: \(\Rightarrow nCu\left(OH\right)2=0,25\left(mol\right)\)
\(Cu\left(OH\right)2\rightarrow CuO+H2O\)
\(\Rightarrow nCuO=0,25\left(mol\right)\Rightarrow mCuO=0,25.80=20\left(g\right)\)
