Question

Answer 1

a, \(n_{CuO}=\dfrac{6}{80}=0,075\left(mol\right)\)

  \(m_{HCl}=250.3,65\%=9,125\left(g\right)\Rightarrow n_{HCl}=\dfrac{9,125}{36,5}=0,25\left(mol\right)\)

PTHH: CuO + 2HCl → CuCl₂ + H₂O

Mol:    0,075     0,15        0,15  

b, Ta có: \(\dfrac{0,075}{1}< \dfrac{0,25}{2}\) ⇒ CuO hết, HCl dư

 \(m_{HCldư}=\left(0,25-0,15\right).36,5=3,65\left(g\right)\)

c, \(m_{CuCl_2}=0,15.135=20,25\left(g\right)\)

d, m_{dd sau pứ} = 6 + 250 = 256 (g)

\(C\%_{ddCuCl_2}=\dfrac{20,25.100\%}{256}=7,91\%\)

\(C\%_{ddHCldư}=\dfrac{3,65.100\%}{256}=1,43\%\)