Fe + 2HCl → FeCl2 + H2
a) \(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)
\(n_{HCl}=\dfrac{24,5}{36,5}=\dfrac{49}{73}\left(mol\right)\)
Theo PT: \(n_{Fe}=\dfrac{1}{2}n_{HCl}\)
Theo bài: \(n_{Fe}=\dfrac{146}{245}n_{HCl}\)
Vì \(\dfrac{146}{245}>\dfrac{1}{2}\) ⇒ Fe dư
Theoo PT: \(n_{Fe}pư=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}\times\dfrac{49}{73}=\dfrac{49}{146}\left(mol\right)\)
\(\Rightarrow n_{Fe}dư=0,4-\dfrac{49}{146}=\dfrac{47}{730}\left(mol\right)\)
\(\Rightarrow m_{Fe}dư=56\times\dfrac{47}{730}=3,61\left(g\right)\)
b) Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}\times\dfrac{49}{73}=\dfrac{49}{146}\left(mol\right)\)
\(\Rightarrow V_{H_2}=\dfrac{49}{146}\times22,4=7,52\left(l\right)\)
