\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
\(nH_2=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
\(nH_2SO_4=nH_2=0,15\left(mol\right)\)
\(nH_2SO_{4\left(15\%\right)}=\dfrac{0,15.15}{100}=0,0225\left(mol\right)\)
\(mH_2SO_4=0,0225.98=2,205\left(g\right)\)
\(nAl=\dfrac{2}{3}.0,15=0,1\left(mol\right)\)
\(mAl=0,1.27=2,7\left(g\right)\)