Ta có: \(n_{Zn}=\dfrac{a}{65}\left(mol\right)\)
\(n_{Ca}=\dfrac{a}{40}\left(mol\right)\)
\(n_{Mg}=\dfrac{a}{24}\left(mol\right)\)
\(n_{Fe}=\dfrac{a}{56}\left(mol\right)\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\) (1)
\(Ca+2HCl\rightarrow CaCl_2+H_2\) (2)
\(Mg+2HCl\rightarrow MgCl_2+H_2\) (3)
\(Fe+2HCl\rightarrow FeCl_2+H_2\) (4)
Theo PT: \(n_{H_2\left(1\right)}=n_{Zn}=\dfrac{a}{65}\left(mol\right)\)
\(n_{H_2\left(2\right)}=n_{Ca}=\dfrac{a}{40}\left(mol\right)\)
\(n_{H_2\left(3\right)}=n_{Mg}=\dfrac{a}{24}\left(mol\right)\)
\(n_{H_2\left(4\right)}=n_{Fe}=\dfrac{a}{56}\left(mol\right)\)
⇒ nH2 (3) lớn nhất → KL pư với HCl dư thu VH2 lớn nhất là Mg.
