a) $2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
b) $n_{Al} = \dfrac{2,7}{27} = 0,1(mol)$
Theo PTHH : $n_{H_2SO_4} = \dfrac{3}{2}n_{Al} = 0,15(mol)$
$m_{H_2SO_4} = 0,15.98 = 14,7(gam)$
c) $n_{H_2} = n_{H_2SO_4} = 0,15(mol) \Rightarrow V_{H_2} = 0,15.22,4 = 3,36(lít)$
d) $C_{M_{H_2SO_4}} = \dfrac{0,15}{0,15} = 1M$