có 2 pt
+, \(2Al+6H_2SO_4->Al_2\left(SO4\right)_3+3SO_2+6H_2O\)
=> \(n_{Al}=\dfrac{n_{SO2}.2}{3}=0,2mol\)
+, \(2Al+4H_2SO_4->Al_2\left(SO_4\right)_3+S+4H2O\)
+> \(n^{_{Al}}=2n_S=2.o,1=0,2mol\)
=> tổng nAl= o,2+0,2=0,4mol
=> mAl=0,4.27=10,8 gam.
Vậy m=10,8gam