$n_{H_2}=6,72/22,4=0,3mol$
$PTHH :$
$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2↑$
Cu không td với H2SO4 loãng
Theo pt :
$n_{Al}=2/3.n_{H_2}=2/3.0,3=0,2mol$
$⇒m_{Al}=0,2.27=5,4g$
$⇒\%m_{Al}=\dfrac{5,4}{10}.100\%=54\%$
$⇒\%m_{Cu}=100\%-54\%=46\%$
b.Theo pt :
$n_{H_2SO_4}=n_{H_2}=0,3mol$
$⇒m_{H_2SO_4}=0,3.98=29,4g$