How many liters of air air is 21 - oxygen are required for the combustion of 6-75 L of CH 4 - Assume air and CH 4 are at the same T and P A- 69-3 LB- 64-3 LC- 72-4 LD- 81-2 L

The correct option is B 64.3 L CH_4 + 2O_2 → 2H_2O + CO_2 Consider PV = nRT. Rewrite it as: n/V = P/R × T Everything on the right side is constant, so th ...

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Standard XII

Chemistry

Avogadro's Law & Gay Lussac's Law

How many lite...

Question

How many liters of air (air is 21% oxygen) are required for the combustion of 6.75 L of CH₄? (Assume air and CH₄ are at the same T and P)

69.3 L

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64.3 L

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72.4L

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81.2L

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Solution

The correct option is B
64.3 L

$C H_{4} + 2 O_{2} \to 2 H_{2} O + C O_{2}$

Consider PV = nRT. Rewrite it as:

$\frac{n}{V} = \frac{P}{R \times T}$

Everything on the right side is constant, so the n:V ratio must also be constant. That means that volume is directly proportional to the number of moles of gas.Since there is a 1:2 molar ratio between $C H_{4}$ and $O_{2}$ .

So,from the given values, we have Volume of $O_{2}$ = 2 $\times$ (6.75 L $C H_{4}$ )

= 13.50L

Or, 21% $O_{2}$ = $\frac{13.50 L o f O_{2}}{x L o f a i r} \times 100$
Or, Volume of air = $\frac{13.50 \times 100}{21}$ = 64.3 L

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How many liters of air (air is 21% oxygen) are required for the combustion of 6.75 L of CH4? (Assume air and CH4 are at the same T and P)

How many liters of air (air is 21% oxygen) are required for the combustion of 6.75 L of CH₄? (Assume air and CH₄ are at the same T and P)