n_{H_2} = (2,688)/(22,4) = 0,12(mol)

n_Y = (0,5376)/(22,4) =0,024(mol)

Đặt n_{Mg} = a(mol); n_{Al} = b(mol)

Gọi khí Y là N_xO_y

Mg + 2HCl -> MgCl_2 + H_2

2Al + 6HCl -> 2AlCl_3 + 3H_2

Theo PT: n_{H_2} = n_{Mg} + 3/2 n_{Al}

=> a + 1,5b = 0,12(1)

(5x-2y)Mg + (12x-4y)HNO_3 -> (5x-2y)Mg(NO_3)_2 + 2N_xO_y + (6x-2y)H_2O

(5x-2y)Al + (18x-6y)HNO_3 -> (5x-2y)Al(NO_3)_3 + 3N_xO_y + (9x-3y)H_2O

Theo PT: n_{N_xO_y} = 2/(5x-2y) . n_{Mg} + 3/(5x-2y) . n_{Al}

=> (2a + 3b)/(5x - 2y) = 0,024

<=> (2.(a+1,5b))/(5x-2y) = 0,024(2)

(1),(2) => (2.0,12)/(5x-2y) = 0,024

<=> (0,24)/(5x-2y) = 0,024

<=> 5x - 2y = 10

=>x = 2; y = 0

=> Y:N_2

n_{H_2}={2,688}/{22,4}=0,12(mol)

Mg^0->Mg^{+2}+2e

Al^0->Al^{+3}+3e

2H^{-1}+2e->H_2^0

Bảo toàn electron: 2n_{Mg}+3n_{Al}=2n_{H_2}=0,24(mol)

n_Y={0,5376}/{22,4}=0,024(mol)

Đặt Y là N_xO_y

Mg^0->Mg^{+2}+2e

Al^0->Al^{+3}+3e

xN^{+5}+(5x-2y).e->N_x^{+2y//x}

Bảo toàn electron: 2n_{Mg}+3n_{Al}=(5x-2y).n_{N_xO_y}

->5x-2y={0,24}/{0,024}=10

->x=2;y=0 là nghiệm duy nhất.

->Y:\ N_2