Khi cho Ag và Al tác dụng với $H_2SO_4$ loãng chỉ có Al phản ứng
$\text{a)}\\2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2\\n_{H_2}=\dfrac{6,72}{22,4}=0,3\,(mol)\\\to n_{Al}=\dfrac{2}{3}n_{H_2}=0,2\,(mol)\\\to m_{Al}=0,2.27=5,4\,(g)\\\to \%Al=\dfrac{5,4}{12,8}.100=42,1875\%\\\to \%Ag=100-42,1875=57,8125\%\\\text{b)}\\n_{Al_2(SO_4)_3}=\dfrac{1}{3}n_{H_2}=0,1\,(mol)\\\to m_{Al_2(SO_4)_3}=0,1.342=34,2\,(g)\\\text{c)}\\n_{H_2SO_4}=n_{H_2}=0,3\,(mol)\\\to V_{H_2SO_4}=\dfrac{0,3}{3}=0,1\,(l)$