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`a)`

Đặt `n_{NO} = x;n_{NO_2} = y`

Ta có pt

`x + y = (4,48)/(22,4) = 0,2(1)`

`M_{khí} = 2 . 19 = 38g//mol`

`=> m_{Khí} = 0,2 . 38 = 7,6g`

`=> 30x + 46y = 7,6(2)`

Từ `(1),(2)`

`=> x = y = 0,1`

`=> n_{NO} = n_{NO_2} = 0,1` mol

`b)`

Đặt `n_{CO_2} = x; n_{O_2} = y`

Ta có pt

`x + y = (5,6)/(22,4) = 0,25(1)`

`M_X = 20 . 2 = 40g//mol`

`=> m_X = 40. 0,25 = 10g`

`=> 44x + 32y = 10(2)`

Từ `(1),(2)`

`=> x = 1/(6); y = 1/(12)`

`=> n_{CO_2} = 1/6` mol; `n_{O_2} = 1/(12)` mol

gọi $\rm x,y$ là số mol $\rm NO,NO_2$

$\rm n_{hh}=\dfrac{4,48}{22,4}=0,2(mol)$

$\rm \to x+y=0,2(1)$

$\rm d_{hh/H_2}=19 \to M=19.2=38(amu)$

$\rm \to \dfrac{30x+46y}{x+y}=38$

`=>`$\rm 30x+46y=38x+38y \to 8x-8y=0(2)$

$\rm (1)(2) \to x=y=0,2$

`---------`

$\rm b.$

gọi $\rm x,y$ là số mol $\rm CO_2,O_2$

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

$\rm \to x+y=0,25(1)$

$\rm d_{hh/H_2}=20 \to M=20.2=40(amu)$

$\rm \to \dfrac{44x+32y}{x+y}=40$

`=>`$\rm 44x+32y=40x+40y \to 4x-8y=0(2)$

$\rm (1)(2) \to x=\dfrac{1}{6};y=\dfrac{1}{12}$