\(n_{H_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\)
PTHH: 2Al + 3H2SO4 ---> Al2(SO4)3 + 3H2
\(\dfrac{7}{15}\)<---0,7<--------------------------0,7
\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=\dfrac{7}{15}.27=12,6\left(g\right)\\m_{H_2SO_4}=0,7.98=68,6\left(g\right)\end{matrix}\right.\)