`m_(dd tăng)=m_(Kl)-m_(H_2)`
`=>m_(H_2)=4,95-4,5=0,45(g)`
`=>n_(H_2)=(0,45)/2=0,225(mol)`
`2Al+3H_2SO_4->Al_2(SO_4)_3+3H_2`
`Mg+H_2SO_4->MgSO_4+H_2`
Đặt `n_(Al)=a;n_(Mg)=b`
`=>27a+24b=4,95`
`3/2*a+b=0,225`
`=>a=0,05;b=0,15`
`=>m_(Al)=27.0,05=1,35(g)`
`m_(Mg)=0,15.24=3,6(g)`
`=>` Chọn `B`