Zn + 2HCl \(\rightarrow\)ZnCl2 + H2
mHCl=\(200.\dfrac{36,5}{100}=73\left(g\right)\)
nHCl=\(\dfrac{73}{36,5}=2\left(mol\right)\)
Theo PTHH ta có:
\(\dfrac{1}{2}\)nHCl=nZn=1(mol)
mZn=65.1=65(g)
b;
Theo PTHH ta có:
nZn=nZnCl2=1(mol)
mZnCl2=136.1=136(g)
c;
Theo PTHH ta có:
nZn=nH2=1(mol)
VH2=22,4.1=22,4(lít)
d;C% dd ZnCl2=\(\dfrac{136}{65+200-1.2}.100\%=51,71\%\)