a,\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PTHH: Zn + 2HCl → ZnCl2 + H2
Mol: 0,1 0,2 0,1 0,1
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b,\(V_{ddHCl}=\dfrac{0,2}{0,2}=1M\)
c,\(C_{M_{ddZnCl_2}}=\dfrac{0,1}{0,2}=0,5M\)
PT: Zn + 2HCl ---> ZnCl2 + H2
nZn = \(\dfrac{6,5}{65}=0,1\left(mol\right)\)
a. Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\)
=> \(V_{H_2}=0,1.22,4=2,24\left(lít\right)\)