a) Gọi $n_{Mg} = x(mol) ; n_{Al} = y(mol) \Rightarrow 24x + 27y = 25,5(1)$

$n_{N_2} = \dfrac{5,6}{22,4} = 0,25(mol)$

Bảo toàn electron : $2n_{Mg} + 3n_{Al} = 10n_{N_2} \Rightarrow 2x + 3y = 0,25.10(2)$

Từ (1)(2) suy ra :  x = 0,5 ; y = 0,5

$m_{Mg} = 0,5.24 = 12(gam)$
$m_{Al} = 0,5.27 = 13,5(gam)$

b) $n_{Mg(OH)_2} = x = 0,5(mol) ; n_{Al(OH)_3} = y = 0,5(mol)$

$\Rightarrow m_{Kết\ tủa} = 0,5.58 + 0,5.78 = 68(gam)$

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3 B

4 D

5 A

6 C

7 B

8 A

9 C

10 B

12 A

12 A

13 D

14 A

15 A

16 C

17 B

18 C

19 A

20 C

21 B

23 C

23 A

24 B

25 B