a,
$n_{HCl}=0,25 mol$
$n_{H_2SO_4}=0,25.0,5=0,125 mol$
$n_{H_2}=0,195 mol$
$Mg+2HCl\to MgCl_2+H_2$
$2Al+6HCl\to 2AlCl_3+3H_2$
$Mg+H_2SO_4\to MgSO_4+H_2$
$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2O$
Ta có $n_H=n_{HCl}+2n_{H_2SO_4}=0,5 mol$
$n_{H\text{pứ}}=0,195.2=0,39<0,5$
Vậy axit dư.
b,
Gọi x, y là mol Mg, Al.
$\Rightarrow 24x+27y=3,87$ (1)
Bảo toàn e: $2x+3y=0,195.2=0,39$ (2)
(1)(2)$\Rightarrow x=0,06; y=0,09$
$\%m_{Mg}=\dfrac{0,06.24.100}{3,87}=39,2\%$
$\%m_{Al}=60,8\%$