2Al + 6HCl -> 2AlCl3 + 3H2
nH2=\(\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PTHH ta có:
\(\dfrac{2}{3}\)nH2=nAl=\(\dfrac{0,2}{3}\left(mol\right)\)
2nH2=nHCl=0,2(mol)
mAl=\(\dfrac{0,2}{3}.27=1,8\left(g\right)\)
C% dd HCl=\(\dfrac{36,5.0,2}{200}.100\%=3,65\%\)