2Al +3S---> Al2S3(1)
Al2S3 +6HCl-----> 2AlCl3 +3H2S(2)
a) Ta có
n\(_{Al}=\frac{0,81}{27}=0,03\left(mol\right)\)
n\(_S=\frac{0,8}{32}=0,025\left(mol\right)\)
=> Al dư
Theo pthh1
n\(_{Al2S3}=\frac{1}{3}n_S=0,083\left(mol\right)\)
Theo pthh2
n\(_{H2S}=3n_{Al2S3}=0,025\left(mol\right)\)
V\(_{H2S}=0,025.22,4=0,56\left(l\right)\)
b) Tự lm nhé
Chúc bạn học tốt