Question

Nhiệt phân hoàn toàn Cu(OH)₂ ra được 8g CuO

a) Viết phương trình phản ứng

b)Tìm khối lượng Cu(OH)₂

Answer 1

a, PTHH: Cu(OH)₂ \(\underrightarrow{t^o}\)CuO + H₂O

b, n_CuO = \(\dfrac{8}{64}=0,125\left(mol\right)\)

Theo PTHH: n_Cu(OH)2 = n_CuO = 0,125 (mol)

m_Cu(OH)2 = 0,125.98 = 12,25 (g)

Answer 2

mk ghi nhầm

b) n_CuO= \(\dfrac{m}{M}\) = \(\dfrac{8}{64}\) = 0,125 mol

\(\Rightarrow\) n\(_{Cu\left(OH\right)_2}\)= 0,125 mol

\(\Rightarrow\) m\(_{Cu\left(OH\right)_2}\)= n.M = 0,125.98 = 12,25g

Answer 3

Cu(OH)₂ \(\underrightarrow{t^o}\) CuO + H₂O

n_CuO = \(\dfrac{8}{64+16}=0,1mol\)

=> \(n_{Cu\left(OH\right)_2}\) = 0,1

=> \(m_{Cu\left(OH\right)_2}\) = 0,1.98 = 9,8g

Answer 4

a) PTHH: Cu(OH)₂ \(\underrightarrow{t^o}\) CuO + H₂O

b) n_CuO= \(\dfrac{m}{M}\) = \(\dfrac{8}{64}\) = 0.125 mol

\(\Rightarrow\) n_{\(Cu\left(OH\right)_2\)}= 0,125 mol

\(\Rightarrow\) m\(_{Cu\left(OH\right)_2}\)= n.M = 0,125.64 = 8g