Bảo toàn nguyên tố:
\(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}.\dfrac{2,7}{27}=0,05\left(mol\right)\)
\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\)
2Al + 6H2SO4 ---> Al2(SO4)3 + 3SO2 + 6H2O
Ta có: nAl = \(\dfrac{2,7}{27}=0,1\left(mol\right)\)
Theo PT: \(n_{Al_2\left(SO_4\right)_3}=0,1.342=34,2\left(g\right)\)
vậy m = 34,2 (g)