a) PTHH: $2Al+3H_2SO_4→Al_2(SO_4)_3+3H_2$
b) $n_{Al}=\dfrac{1,35}{27}=0,05(mol)$
$n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}×0,05=0,075(mol)$
→ $V_{H_2}=0,075×22,4=1,68(l)$
c) $n_{H_2SO_4}=n_{H_2}=0,075(mol)$
→ $m_{H_2SO_4}=0,075×98=7,35(g)$
→ $C$%$_{ddH_2SO_4}=\dfrac{7,35}{50}×100$% $=$ $14,7$%