Ta co pthh
Zn + 2HCl \(\rightarrow\) ZnCl2 + H2
Theo de bai ta co
nH2=\(\dfrac{5,6}{22,4}=0,25mol\)
Theo pthh
nZn=nH2=0,25 mol
\(\Rightarrow mZn=0,25.65=16,25g\)
Vay khoi luong kem can dung la 16,25 g
nH2=V/22,4=5,6/22,4=0,25(mol)
PT:
Zn + 2HCl -> ZnCl2 + H2
1.........2............1............1 (mol)
0,25 < -0,5 <-0,25< - 0,25 (mol)
=> \(m_{Zn}=n.m=0,25.65=16,25\left(g\right)\)