a. 2Al+3H2SO4\(\rightarrow\)Al2(SO4)3+3H2
b. nH2=\(\frac{13,44}{22,4}\)=0,6
\(\rightarrow\)nAl=nH2.\(\frac{2}{3}\)=0,6.\(\frac{2}{3}\)=0,4
\(\rightarrow\)mAl=0,4.27=10,8
\(\rightarrow\)mCu=13-10,8=2,2
c. %mAl=\(\frac{10,8}{13}\)=83,07%
\(\rightarrow\)%mCu=100%-83,07%=16,93%