Zn + 2HCl → ZnCl2 + H2
\(n_{Zn}=\frac{13}{65}=0,2\left(mol\right)\)
\(m_{HCl}=365\times5\%=18,25\left(g\right)\)
\(\Rightarrow n_{HCl}=\frac{18,25}{36,5}=0,5\left(mol\right)\)
Theo PT: \(n_{Zn}=\frac{1}{2}n_{HCl}\)
Theo bài: \(n_{Zn}=\frac{2}{5}n_{HCl}\)
Vì \(\frac{2}{5}< \frac{1}{2}\) ⇒ HCl dư, tính theo Zn
a) Theo pT: \(n_{H_2}=n_{Zn}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2\times22,4=4,48\left(l\right)\)
b) \(m_{H_2}=0,2\times2=0,4\left(g\right)\)
Theo pT: \(n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\)
\(\Rightarrow m_{ZnCl_2}=0,2\times136=27,2\left(g\right)\)
c) DD sau pư gồm: HCl dư và ZnCl2
Ta có: \(m_{dd}saupư=13+365-0,4=377,6\left(g\right)\)
\(C\%_{ZnCl_2}=\frac{27,2}{377,6}\times100\%=7,2\%\)
Theo pT: \(n_{HCl}pư=2n_{Zn}=2\times0,2=0,4\left(mol\right)\)
\(\Rightarrow n_{HCl}dư=0,5-0,4=0,1\left(mol\right)\)
\(\Rightarrow m_{HCl}dư=0,1\times36,5=3,65\left(g\right)\)
\(C\%_{HCl}dư=\frac{3,65}{377,6}\times100\%=0,97\%\)
