Xin câu trả lời hay nhất ạ!
Ta có: $n_{Al}$ =$\frac{5,4}{27}$ =$0,2(mol)$
a) pt : $2Al$ + $6H_{2}SO_{4}$→$Al_{2}(SO_{4})_{3}$ + $3SO_{2}$ + $6H_{2}O$
b)Theo pt có: $n_{SO_{2}}$ = $\frac{3}{2}$$n_{Al}$ = $\frac{3}{2}$$×0,2=0,3(mol)$
⇒$V_{SO_{2}}$ =$0,3×22,4=6,72(l)$
