2K + 2HCl ➜ 2KCl + H2
\(n_K=\dfrac{7,8}{39}=0,2\left(mol\right)\)
\(m_{HCl}=100\times36,5\%=36,5\left(g\right)\)
\(\Rightarrow n_{HCl}=\dfrac{36,5}{36,5}=1\left(mol\right)\)
Theo PT: nK = nHCl
Theo bài: \(n_K=\dfrac{1}{5}n_{HCl}\)
Vì \(\dfrac{1}{5}< 1\) ⇒ Kali hết, dd HCl dư
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_K=\dfrac{1}{2}\times0,2=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,1\times22,4=2,24\left(l\right)\)
Dung dịch A gồm: KCl và HCl dư
Theo PT: \(n_{KCl}=n_K=0,2\left(mol\right)\)
\(\Rightarrow m_{KCl}=0,2\times74,5=14,9\left(g\right)\)
Theo PT: \(n_{HCl}pư=n_K=0,2\left(mol\right)\)
\(\Rightarrow n_{HCl}dư=1-0,2=0,8\left(mol\right)\)
\(\Rightarrow m_{HCl}dư=0,8\times36,5=29,2\left(g\right)\)
\(\Sigma m_{dd}=7,8+100=107,8\left(g\right)\)
\(C\%_{ddKCl}=\dfrac{14,9}{107,8}\times100\%=13,82\%\)
\(C\%_{ddHCl}dư=\dfrac{29,2}{107,8}\times100\%=27,09\%\)
