Theo đề ta có : nCu(OH)2 = \(\dfrac{19,6}{98}=0,2\left(mol\right)\)
PTHH :
Cu(OH)2 \(-^{t0}->CuO+H2O\)
0,2mol.....................0,2mol
CuO + H2\(-^{t0}->Cu+H2O\)
0,2mol...................0,2mol
=> mCu = 64.0,2=12,8(g)
Cu(OH)2 ➜ CuO + H2O (1)
CuO + H2 ➜ Cu + H2O (2)
\(n_{Cu\left(OH\right)_2}=\dfrac{19,6}{98}=0,2\left(mol\right)\)
Theo PT1: \(n_{CuO}=n_{Cu\left(OH\right)_2}=0,2\left(mol\right)\)
Số mol CuO ở PT1 chính là số mol CuO ở PT2
Theo PT2: \(n_{Cu}=n_{CuO}=0,2\left(mol\right)\)
⇒ \(m_{Cu}=0,2.64=12,8\left(g\right)\)
nCu(OH)2 = 19.6/98=0.2mol
Cu(OH)2 -> CuO + H2O
(mol) 0.2 0.2
CuO + H2 -> Cu + H2O
(mol) 0.2 0.2
mCu = 0.2*64 = 12.8g