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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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iliketorifk iliketorifk
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12 years ago
1) How many litres of oxygen are necessary for the combustion of 277g of carbon monoxide assuming that the reaction occurs at STP?  The balanced equation is:
2CO+O2 -> 2CO2


2)Propane (C3H8) burns oxygen to produce carbon dioxide and water vapor.  the balanced equation for this reaction is C3H8+5O2->4H2O+3CO2. What volume of carbon dioxide is produced when 2.8L of oxygen are consumed?
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#1
12 years ago
first off the phraseology is incorrect.  FYI   Carbon monoxide does not burn(combust).  It is a resultant gas produced from combustion.
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tonton
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#2
Posts: 102
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12 years ago
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#3
12 years ago
#1
First, convert the 277g of CO to moles of CO.

277g CO (1mole CO/28 g CO) = 9.89 moles CO

Then, using the balanced equation divide the number of moles of CO by 2.  This answer will be the number of moles of O2 required to consume the moles of CO. The reason you divide by 2 is that each mole of CO only requires 1/2 mole of O2 (from the balanced equation).

9.89 Moles/2 = 4.95 moles of O2 consumed.

Finally, multiply the number of moles of O2 by 22.4 L/mole. You use 22.4 because you are at STP. 22.4 is called a "molar volume" (at STP).

Volume of O2 = 22.4 L/mole x 4.95 moles O2

Volume of O2 = 111 liters of O2


#2

You use the coefficients of the balanced equation:

For each liter of O2 consumed, 3 liters of CO2 are produced.
For gases you can use these coeffients as you would moles. "Equal volumes of gases measured under the same conditions, have equal numbers of molecules".

Liters of CO2 produced = 2.8 liters O2(3 liter CO2/5 liter O2 )

Liters of CO2 produced = 1.7 liters CO2

Hope this helps!
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