3Mg + 8HNO3 -> 3Mg(NO3)2 + 2NO + 4H2O

    a                                                2a/3

Al + 4HNO3 -> Al(NO3)3 + NO + 2H2O

 b                                           b

=> 24a + 27b = 2.7

2a/3 + b = 0.08

=> a = 0.09 và b = 0.02

% mMg = 0.09 x 24/2.7 = 80%

=> % mAl = 20%

nHNO3 = 8a/3 + 4b = 0.32 mol

=> CM HNO3 = 0.32/0.1 = 3.2 M

trong dd X chứa Mg(NO3)2: 0.09 mol và Al(NO3)3: 0.02 mol

=> CM Mg(NO3)2 = 0.09/0.1 = 0.9 M

CM Al(NO3)3 =  0.02/0.1 = 0.2 M

Bạn tham khảo:

$3Mg+8HNO_3 \to 3Mg(NO_3)_2+2NO+4H_2O$

$Al+4HNO_3 \to Al(NO_3)_3+NO+2H_2O$

$a/$

$n_{Mg}=a(mol)$

$n_{Al}=b(mol)$

$m_{hh}=24a+27b=2,7(1)$

$n_{NO}=0,08(mol)$

$n_{NO}=\frac{2}{3}.a+b=0,08(2)$

$a=0,09; b=0,02$
$\%m_{Mg}=\frac{0,09.24}{2,7}.100\%=80\%$

$\%m_{Al}=20\%$

$b/$

$n_{HNO_3}=\frac{8}{3}a+4b=\frac{8}{3}.0,09+0,02.4=0,32(mol)$

$CM=\frac{0,32}{0,1}=3,2M$

$c/$

$CM_{Mg(NO_3)_2}=\frac{0,09}{0,1}=0,9M$

$CM_{Al(NO_3)_3}=\frac{0,02}{0,1}=0,2M$