Posted by Lance on Friday, May 4, 2012 at 1:49am.
This is an example of a combustion reaction involving a hydrocarbon (in this case propane) reacting with oxygen. The products of such reactions are almost always water and carbon dioxide.
C3H8(g) + O2(g) → H2O + CO2
First balance the number of Carbons on both sides. There are 3 Carbon atoms for every propane molecule and 1 for each carbon dioxide.
C(3)H8(g) + O2(g) → H2O + _X_CO2
Second balance the number of Hydrogens. Here we have 8 for each propane and 2 for each water.
C3H(8)(g) + O2(g) → _Y_H(2)O + X*CO2
Finally count up all the oxygen atoms on the right side of the equation and balance that with the oxygen molecules on the left.
C3H8(g) + _Z_O2(g) → Y*H2O + X*CO2
where Z = X/2 + Y
Sometimes after this you will end up with fractions, which is okay, but if you prefer you can multiply both sides of the equation by the lowest common denominator to get rid of the fractions. In this case you shouldn't have any fractions.
Oh, and don't forget to include the states. CO2 is obviously a gas and H2O will likely also be a gas (water vapor) due to the heat yielded by the reaction.
Hope this helps.
2[C3H8]+7O2 - 3[CO2](g)+8[H2O](l)
2[C3H8]+7O2 - 3[CO2](g)+8[H2O](l)
2[C3H8] +7O2 - 3[CO2](g) + 8[H2O](l)
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