Cu(OH)2 \(\underrightarrow{to}\) CuO + H2O
\(n_{Cu\left(OH\right)_2}=\frac{14,7}{98}=0,15\left(mol\right)\)
Theo pT: \(n_{CuO}=n_{Cu\left(OH\right)_2}=0,15\left(mol\right)\)
\(\Rightarrow m_{CuO}=0,15\times80=12\left(g\right)\)
\(n_{Cu\left(OH\right)_2}=\frac{14,7}{98}=0,15mol\)
PTHH :\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\) (1)
(mol ) 0,15 \(\rightarrow\) 0,15
Từ (1) => \(n_{CuO}=0,15mol\Rightarrow m_{CuO}=0,15.80=12g\)
\(n_{Cu\left(OH\right)_2}=\frac{14,7}{98}=0,15\left(mol\right)\)
PTHH: \(Cu\left(OH\right)_2\underrightarrow{t^0}CuO+H_2O\)
Theo PTHH: \(n_{Cu\left(OH\right)_2}:n_{CuO}=1:1\)
\(\Rightarrow n_{CuO}=n_{Cu\left(OH\right)_2}=0,15\left(mol\right)\)
\(\Rightarrow m_{CuO}=0,15.80=12\left(g\right)\)
nCu(OH)2 = 14.7/98=0.15 mol
Cu(OH)2 -to-> CuO + H2O
0.15__________0.15
mCuO= 0.15*80 = 12 g
nCu(OH)2 = 14.7/98=0.15 mol
Cu(OH)2 ----(to)----> CuO + H2O
0.15.............................0.15.....................(mol)
mCuO= 0.15*80 = 12 (g)
Cu(OH)2 ----to--> CuO + H2O
nCuO = 14,7 / 98=0,15 (mol)
Theo PTTT: nCuO = nCu(OH)2 =0,15 (mol)
m CuO = 0,15*80=12 (g)
