a) Zn + H2SO4 → ZnSO4 + H2
\(n_{Zn}=\frac{1,95}{65}=0,03\left(mol\right)\)
\(n_{H_2SO_4}=\frac{1,47}{98}=0,015\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2SO_4}\)
Theo bài: \(n_{Zn}=2n_{H_2SO_4}\)
Vì \(2>1\) ⇒ Zn dư
b) Theo PT: \(n_{Zn}pư=n_{H_2SO_4}=0,015\left(mol\right)\)
\(\Rightarrow n_{Zn}dư=0,03-0,015=0,015\left(mol\right)\)
\(\Rightarrow m_{Zn}dư=0,015\times65=0,975\left(g\right)\)
c) Theo PT: \(n_{H_2}=n_{H_2SO_4}=0,015\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,015\times22,4=0,336\left(l\right)\)