a, nZn = \(\frac{6,5}{65}=0,1mol\)
Đổi 100ml = 0,1l
pt : Zn + 2HCl → ZnCl2 + H2
(mol) 0,1mol → 0,1mol → 0,1mol
b, VH2 = 0,1 .22,4 = 2,24l
c, CM = \(\frac{0,1}{\frac{100}{1000}}\) = 1M
a) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b) Ta có: \(n_{Zn}=\frac{6,5}{65}=0,1\left(mol\right)\) \(\Rightarrow n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\)
c) Theo PTHH: \(n_{Zn}=n_{HCl}=0,1\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\frac{0,1}{0,1}=1\left(M\right)\)