* Tóm tắt:
Biết: mZn = 6,5 (g)
CM(dd.HCl) = 0,5M
Hỏi: a) VH2 = ?
b) Vdd HCl = ?
a) \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,1-->0,2------------->0,1
VH2 = 0,1.22,4 = 2,24 (l)
b) \(V_{\left(dd.HCl\right)}=\dfrac{0,2}{0,5}=0,4\left(l\right)\)
Tóm tắt:
`m_[Zn]=6,5 g`
`C_[M_[HCl]]=0,5 M`
________________________
`a)V_[H_2]=? l`
`b)V_[dd HCl]= ? l`
Giải:
`Zn + 2HCl -> ZnCl_2 + H_2`
`0,1` `0,2` `0,1` `(mol)`
`n_[Zn]=[6,5]/65=0,1(mol)`
`a)V_[H_2]=0,1.22,4=2,24(l)`
`b)V_[dd HCl]=[0,2]/[0,5]=0,4(l)`