\(n_{Zn}=\dfrac{2,6}{65}=0,04\left(mol\right)\)
PTHH: Zn + 2HCl → ZnCl2 + H2
Mol: 0,04 0,08 0,04
\(\Rightarrow C_{M_{ddHCl}}=\dfrac{0,08}{0,2}=0,4M\)
\(V_{H_2}=0,04.22,4=0,896\left(l\right)\)
Đúng 2
Bình luận (0)
PTHH: Zn + 2HCl → ZnCl2 + H2
\(\Rightarrow n_{HCl}=2n_{Zn}=2.\dfrac{2,6}{65}=0,08\left(mol\right)\)
\(\Rightarrow C_{M\left(HCl\right)}=\dfrac{0,08}{0,2}=0,4M\)
\(n_{H_2}=n_{Zn}=0,08\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,08.22,4=1,792\left(l\right)\)
Đúng 0
Bình luận (2)