2Al + 6HCl \(\rightarrow\)2AlCl3 + 3H2
nAl=\(\dfrac{27}{27}=1\left(mol\right)\)
nHCl=\(\dfrac{14,6}{36,5}=0,4\left(mol\right)\)
Vì \(\dfrac{0,4}{3}< 1\)nên Al dư 1-\(\dfrac{0,4}{3}\)=\(\dfrac{2,6}{3}\)(mol)
mAl dư=27.\(\dfrac{2,6}{3}\)=23,4(g)
Theo PTHH ta có:
\(\dfrac{1}{2}\)nHCl=nH2=0,2(mol)
VH2=0,2.22,4=4.48(lít)