1) $Zn + 2HCl \to ZnCl_2 + H_2$
2) $n_{H_2} = n_{Zn} = \dfrac{13}{65} = 0,2(mol)$
$V_{H_2} = 0,2.22,4 = 4,48(lít)$
1. PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
2. \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
\(\Rightarrow n_{Zn}=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)