Zn + 2 HCl \(\rightarrow\)ZnCl2 + H2 (1)
nZn=\(\dfrac{19,5}{65}=0,3\left(mol\right)\)
a;
Theo PTHH 1 ta có:
nZn=nH2=0,3(mol)
VH2=0,3.22,4=6,72(lít)
b;
Fe2O3+ 3H2 \(\rightarrow\)2Fe + 3H2O (2)
nFe2O3=\(\dfrac{19,2}{160}=0,12\left(mol\right)\)
Vì \(\dfrac{0,3}{3}< 0,12\)nên Fe2O3 dư
Theo PTHH 2 ta có:
nFe=\(\dfrac{2}{3}\)nH2=0,2(mol)
mFe=56.0,2=11,2(g)