PTHH: Zn + H2SO4 → ZnSO4 + H2
\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)
\(n_{H_2SO_4}=0,25\times2,4=0,6\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2SO_4}\)
Theo bài: \(n_{Zn}=\dfrac{1}{2}n_{H_2SO_4}\)
Vì \(\dfrac{1}{2}< 1\) ⇒ HCl dư
Theo PT: \(n_{H_2}=n_{Zn}=0,3\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,3\times22,4=6,72\left(l\right)\)
Vì \(HS=90\%\) \(\Rightarrow V_{H_2}tt=6,72\times90\%=6,048\left(l\right)\)