\(\begin{array}{l}
a)\\
2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\
n_{Al}=\frac{5,4}{27}=0,2(mol)\\
n_{Al_2(SO_4)_3}=\frac{1}{2}n_{Al}=0,1(mol)\\
m_{Al_2(SO_4)_3}=0,1.342=34,2(g)\\
b)\\
n_{H_2}=\frac{3}{2}n_{Al}=0,15(mol)\\
V_{H_2}=0,15.24,79=3,7185(l)\\
c)\\
n_{H_2SO_4}=n_{H_2}=0,15(mol)\\
m_{dd\,H_2SO_4}=\frac{0,15.98}{9,8\%}=150(g)\\
d)\\
C\%_{Al_2(SO_4)_3}=\frac{34,2}{5,4+150-0,15.2}.100\%\approx 22,05\%
\end{array}\)