Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: LoveInQuanta on September 21, 2009, 06:16:43 PM
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Howdy, all you fine folks.
I'm in Accelerated General Chem, and I have encountered the problem copied below.
""Strike anywhere" matches contain the compound tetraphosphorus trisulfide, which burns to form tetraphosphorus decaoxide and sulfur dioxide gas. How many milliliters of sulfur dioxide, measured at 732 torr and 32°C, can be produced from burning 0.841 g of tetraphosphorus trisulfide?"
I believe I can field the problem if I could just write and balance the equation. What I have so far is:
P4S3 --> P4O10 + SO2(g)
...but obviously the oxygen is not present on the reactant side of the equation. Where does it appear? Is it a reactant as O2 gas which is used to burn the P4S3?
Many thanks!
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Air oxygen.
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Any time you see "burn", "combustion," or something of the like, then oxygen is present. You can't have a fire without O2.
Some people leave it out of the equation because it's a given in any combustion reaction. It's not particularly difficult to balance out on the other side if it helps, but it's not necessary if the problem states what was done in order to make the reaction happen (ie, burning).
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Thanks, Borek & Tin Man. It may be a given, but it's not for those of us who are utterly new to this. :)
Problem solved.
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Gasses as a percentage of clean dry air (one estimate)
Nitrogen 78.084
Oxygen 20.947
Argon 0.934
Carbon Dioxide 0.033
other 0.002