2Al +6HCl ----->2AlCl3 +3H2
Ta có
n\(_{Al}=\frac{8,1}{27}=0,3\left(mol\right)\)
n\(_{HCl}=0,2.1=0,2\left(mol\right)\)
=> Al dư
Theo pthh
n\(_{H2}=\frac{1}{2}n_{HCl}=0,1\left(mol\right)\)
V\(_{H2}=01.22,4=2,24\left(l\right)\)
Theo pthh
n\(_{AlCl3}=\frac{1}{3}n_{HCl}=0,067\left(mol\right)\)
m\(_{AlCl3}=0,067.133,5=8,9445\left(g\right)\)
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