Phương trình hóa học:
Zn + 2HCl => ZnCl2 + H2
nZn = m/M = 6.5/65 = 0.1 (mol)
Theo phương trình => nH2 = 0.1 (mol), nHCl = 0.2 (mol)
VH2 = 22.4 x 0.1 = 2.24 (l)
VddHCl = n/CM = 0.2/1 = 0.2 (l) = 200ml
a) PTHH: Zn + 2HCl \(\rightarrow\) ZnCl2 + H2\(\uparrow\)
nZn =\(\frac{6,5}{65}=0,1\left(mol\right)\)
Theo PT: n\(H_2\) = nZn = 0,1 (mol)
=> V\(H_2\) = 0,1.22,4 = 2,24 (l)
b) Theo PT: nHCl = 2nZn = 2.0,1 = 0,2 (mol)
=> VHCl = \(\frac{0,2}{1}=0,2\left(l\right)=200\left(ml\right)\)
nZn= 0.1 mol
Zn + 2HCl --> ZnCl2 + H2
0.1 ___0.2___________0.1
VH2= 0.1*22.4= 2.24l
VddHCl= 0.2/ 1= 0.2l