30 ml gaseous mixture of methane and ethylene in volume ratio X : Y requires 350 ml air containing 20% of O 2 by volume for complete combustion. If ratio of methane and ethylene changed to Y : X. Wh

Question

30 ml gaseous mixture of methane and ethylene in volume ratio X : Y requires 350 ml air containing 20% of O 2 ​ by volume for complete combustion. If ratio of methane and ethylene changed to Y : X. What will be volume of air (in ml) required for complete reaction under similar condition of temperature and pressure. [Divide Answer by 100]

TestHub · Accepted Answer

For combustion: CH₄ + 2O₂ → CO₂ + 2H₂O

C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

Let CH₄ = 30X and C₂H₄ = 30Y, where $X + Y = 1$ .

Air used = 350 mL, so O₂ = $350 \times 20% = 70$ mL.

Therefore,

$30 (2 X + 3 Y) = 70 \Rightarrow 2 X + 3 Y = \frac{7}{3}$

With $X + Y = 1$ , we get $X = \frac{2}{3}$ , $Y = \frac{1}{3}$ . On reversing the ratio, CH₄:C₂H₄ = 1:2, so CH₄ = 10 mL, C₂H₄ = 20 mL.

O₂ required = $10 (2) + 20 (3) = 80$ mL. Hence air required:

$\frac{80}{0.20} = 400 m L$ Given divide answer by 100:

$4$

Answer = $4$

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