$a,PTPƯ:$
$Mg+H_2SO_4\xrightarrow{} MgSO_4+H_2↑$ $(1)$
$2Al+3H_2SO_4\xrightarrow{} Al_2(SO_4)_3+3H_2↑$ $(2)$
Gọi $n_{Mg}$ là a (mol), $n_{Al}$ là b (mol).
$⇒24a+27b=7,8$ $(*)$
$n_{H_2}=\dfrac{8,96}{22,4}=0,4mol.$
$Theo$ $pt1:$ $n_{H_2}=n_{Mg}$
$Theo$ $pt2:$ $n_{H_2}=1,5n_{Al}$
$⇒a+1,5b=0,4.$ $(**)$
$(*)$ và $(**)$ :
$\left \{ {{a+1,5b=0,4} \atop {24a+27b=7,8}} \right.$ $⇒\left \{ {{a=0,1} \atop {b=0,2}} \right.$
$⇒\%m_{Mg}=\dfrac{0,1.24}{7,8}.100\%=30,77\%$
$⇒\%m_{Al}=\dfrac{0,2.27}{7,8}.100\%=69,23\%$
$b,Theo$ $pt1:$ $n_{H_2SO_4}=n_{Mg}=0,1mol.$
$Theo$ $pt2:$ $n_{H_2SO_4}=\dfrac{3}{2}n_{Al}=0,3mol.$
$⇒C\%_{H_2SO_4}=\dfrac{(0,1+0,3).98}{200}.100\%=19,6\%$
chúc bạn học tốt!