Gọi nAl=a, nMg=b

nO2=8,96/22,4=0,4mol

4Al+3O2→2Al2O3

a      0,75a    0,5a

2Mg+O2→2MgO

b       0,5b    b

27a+24b=15,6

0,75a+0,5b=0,4

→a=0,4

   b=0,2

mAl=0,4.27=10,8g

%mAl=10,8/15,6.100%=69,23%

%mMg=100-69,23=30,77%

mAl2O3+mMgO=0,5.0,4.102+0,2.40=28,4g

$a)$ Gọi $\left \{ {{n_{Al}=x(mol)} \atop {n_{Mg}=y(mol)}} \right.$ 

$m_{hh}=27x+24y=15,6(g)(1)$

$n_{O_{2}}=\frac{8,96}{22,4}=0,4(mol)$

$PTHH:4Al+3O_{2} \xrightarrow {t^o} 2Al_{2}O_{3}$

$mol:$ $x$ →$0,75x$               

$PTHH:2Mg+O_{2} \xrightarrow {t^o} 2MgO$

$mol:$ $y$   →$0,5y$

$n_{O_{2}}=0,75x+0,5y=0,4(mol)(2)$

 Giải hệ $(1)$,$(2)$⇒$\left \{ {{x=0,4} \atop {y=0,2}} \right.$ 

%$m_{Al_{2}O_{3}}=\frac{0,4.27}{15,6}.100$%$=69,23$%

%$m_{MgO}=100$%$-69,23$%$=30,77$%

$b)$ Theo BTKL: $m_{oxit}=m_{hh kim loại}+m_{O_{2}}=15,6+0,4.32=28,4(g)$