Zn + 2HCl → ZnCl2 + H2 (1)
H2 + CuO \(\underrightarrow{to}\) Cu + H2O (2)
\(n_{Zn}=\dfrac{3}{65}\left(mol\right)\)
\(m_{ddHCl}=18,96\times1,07=20,2872\left(g\right)\)
\(\Rightarrow m_{HCl}=20,2872\times14,6\%=2,9619312\left(g\right)\)
\(\Rightarrow n_{HCl}=\dfrac{2,9619312}{36,5}=0,0811488\left(mol\right)\)
Theo PT1: \(n_{Zn}=\dfrac{1}{2}n_{HCl}\)
Theo bài: \(n_{Zn}=0,5688n_{HCl}\)
Vì \(0,5688>\dfrac{1}{2}\) ⇒ Zn dư
Theo PT1: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}\times0,0811488=0,0405744\left(mol\right)\)
theo PT2: \(n_{Cu}=n_{H_2}=0,0405744\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,0405744\times64=2,5967616\left(g\right)\)
