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What volume of oxygen is necessary to completely react with 125 ml of methane gas?
CH4 + O2 -> CO2 + H2O
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3 Answers
- pisgahchemistLv 78 years agoFavorite Answer
With all due respect, Bee Killer has forgotten about Gay-Lussac's law of combining volumes. The volumes of the gases in the reaction will be in the same ratio as the moles, so there is no need to convert to moles.
Specifically: "The law of combining volumes states that, when gases react together to form other gases, and all volumes are measured at the same temperature and pressure: The ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers."
http://en.wikipedia.org/wiki/Gay-Lussac%27s_law (This by the way, this is the ONLY "Gay-Lussac's law.")
You must start from a balanced chemical equation...
CH4 + 2O2 -> CO2 + 2H2O
125mL ??mL
125 mL CH4 x (2 mL O2 / 1 mL CH4) = 250. mL O2 ........ to three significant digits
============= Follow up =============
In addition, do not round off the intermediate answer, but use it to calculate the moles of O2, Bee Killer wrote it as CH4. He made a mistake in converting from moles to the volume of O2 at STP. He used the atomic weight of oxygen atoms instead of 22.4 L/mol. All of that could have been alleviated by using Gay-Lussac's law of combining volumes.
- bee killerLv 78 years ago
First you have to balance the equation
CH4 + 2O2 -> CO2 + 2H2O
The mole ratio of methane to oxygen is 1:2
Change ml of gas to liters. = .125liters
.125/22.4 = .00558 moles of methane
The mole ratio shows that you need twice as much of the oxygen to the methane = .011 moles of CH4
.011 moles of CH4 = .18 Liters of O2
- ?Lv 44 years ago
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