Zn + 2HCl \(\rightarrow\)ZnCl2 + H2
nZn=\(\dfrac{13}{65}=0,2\left(mol\right)\)
Theo PTHH ta có:
nZn=nH2=0,2(mol)
VH2=0,2.22,4=4,48(lít)
c;Theo PTHH ta có:
nZn=nZnCl2=0,2(mol)
mZnCl2=136.0,2=27,2(g)
mdd HCl=200.1,2=240(g)
C% dd ZnCl2=\(\dfrac{27,2}{240+13-0,2.2}.100\%=10,768\%\)