2Al + 6HCl \(\rightarrow\)2AlCl3 + 3H2 (1)
Al2O3 + 6HCl \(\rightarrow\)2AlCl3 + 3H2O (2)
nH2=\(\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
Theo PTHH 1 ta có:
\(\dfrac{2}{3}\)nH2=nAl=\(\dfrac{0,1}{3}\left(mol\right)\)
mAl=\(\dfrac{0,1}{3}.27=0,9\left(g\right)\)
mAl2O3=16,5-0,9=15,6(g)
b;
nAl2O3=\(\dfrac{15,6}{102}=0,153\left(mol\right)\)
Theo PTHH 1 và 2 ta cso:
3nAl=nHCl=0,15(mol)
6nAl2O3=nHCl=0,918(mol)
mHCl=36,5.(0,15+0,918)=38,982(g)
mdd HCl=38,982:14,6%=267g