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)

A

69.3 L

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B

64.3 L

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C

72.4L

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D

81.2L

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Solution

## The correct option is B 64.3 L CH4+2O2→2H2O+CO2 Consider PV = nRT. Rewrite it as: nV=PR×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 CH4 and O2. So,from the given values, we have Volume of O2 = 2×(6.75 L CH4) = 13.50L Or, 21% O2 = 13.50L of O2xL of air×100 Or, Volume of air = 13.50×10021 = 64.3 L

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