Question

Nung hoàn toàn 14,7 g Cu(OH)₂ thu được chất rắn là CuO và hơi nước.Tính khối lượng của CuO thu được ?

Answer 1

Cu(OH)₂ \(\underrightarrow{to}\) CuO + H₂O

\(n_{Cu\left(OH\right)_2}=\frac{14,7}{98}=0,15\left(mol\right)\)

Theo pT: \(n_{CuO}=n_{Cu\left(OH\right)_2}=0,15\left(mol\right)\)

\(\Rightarrow m_{CuO}=0,15\times80=12\left(g\right)\)

Answer 2

\(n_{Cu\left(OH\right)_2}=\frac{14,7}{98}=0,15mol\)

PTHH :\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\) (1)

(mol ) 0,15 \(\rightarrow\) 0,15

Từ (1) => \(n_{CuO}=0,15mol\Rightarrow m_{CuO}=0,15.80=12g\)

Answer 3

\(n_{Cu\left(OH\right)_2}=\frac{14,7}{98}=0,15\left(mol\right)\)

PTHH: \(Cu\left(OH\right)_2\underrightarrow{t^0}CuO+H_2O\)

Theo PTHH: \(n_{Cu\left(OH\right)_2}:n_{CuO}=1:1\)

\(\Rightarrow n_{CuO}=n_{Cu\left(OH\right)_2}=0,15\left(mol\right)\)

\(\Rightarrow m_{CuO}=0,15.80=12\left(g\right)\)

Answer 4

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

Answer 5

n_Cu(OH)2 = 14.7/98=0.15 mol

Cu(OH)2 ----(t^o)----> CuO + H2O

0.15.............................0.15.....................(mol)

m_CuO= 0.15*80 = 12 (g)

Answer 6

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)