a)

$n_{CO} = \frac{1,2395}{24,79} = 0,05 \text{ (mol)}$

$n_{CO_2} = \frac{3,7185}{24,79} = 0,15 \text{ (mol)}$

$\Rightarrow m_A = (0,05 \cdot 28) + (0,15 \cdot 44) = 8 \text{ (g)}$

b)

$n_A = 0,05 + 0,15 = 0,2 \text{ (mol)}$

$\overline{M}_A = \frac{8}{0,2} = 40 \text{ (g/mol)}$

$\Rightarrow d_{A/H_2} = \frac{40}{2} = 20$

c)

$\overline{M}_B = 20,4 \cdot 2 = 40,8 \text{ (g/mol)}$

$\text{Sử dụng quy tắc đường chéo:}$

$\frac{V_{CO}}{V_{CO_2}} = \frac{44 - 40,8}{40,8 - 28} = \frac{3,2}{12,8} = \frac{1}{4}$

$\text{Vậy tỉ lệ thể tích cần trộn là 1 : 4.}$

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