a) PTHH: 2Al + 3H2SO4 \(\rightarrow\) Al2(SO4)3 + 3H2
Ta có: nAl = \(\dfrac{5,4}{27}=0,2\left(mol\right)\)
Theo phương trình, nH2 = \(\dfrac{0,2\times3}{2}=0,3\left(mol\right)\)
=> VH2(đktc) = \(0,3\cdot22,4=6,72\left(l\right)\)
b) Theo phương trình, nAl2(SO4)3 = \(\dfrac{0,2}{2}=0,1\left(mol\right)\)
=> mAl2(SO4)3 = \(0,1\cdot342=34,2\left(gam\right)\)