$2Al + 3H_2SO_4 → Al_2(SO_4)_3 + 3H_2$
$a\hspace{5cm}1,5a$
$Mg + H_2SO_4 → MgSO_4 + H_2$
$b\hspace{5cm}b$
$m_{Al, Mg}=5,1 (g)$
$⇒27a+24b=5,1\quad (1)$
$n_{H_2}=\frac{5,6}{22,4}=0,25 (mol)$
$⇒1,5a+b=0,25 \quad(2)$
Từ (1), (2) $⇒\left\{{{27a+24b=5,1}\atop{1,5a+b=0,25}}\right.$
$⇔\left\{{{27a+24b=5,1}\atop{b=0,25-1,5a}}\right.$
$⇔\left\{{{27a+24(0,25-1,5a)=5,1}\atop{b=0,25-1,5a}}\right.$
$⇔\left\{{{a=0,1}\atop{b=0,25-0,1.1,5=0,1}}\right.$
$⇒m_{Al}=0,1.27=2,7 (g)$
$⇒m_{Mg}=5,1-2,7=2,4 (g)$