Balance the combustion reaction between butane and oxygen. 2C^4H^10+______O^2→___CO^2+____H^2O
How many carbon atoms can you see on the reactant-side?
Combustion yields water and carbon dioxide from the butane and oxygen gas. C4H10 + ?O2 = ?CO2 + ?H2O One molecule of butane has 10 hydrogens, so it will yield 5 molecules of water with 2 hydrogens each. With 4 carbons, it can yield 4 molecules of carbon dioxide. C4H10 + ?O2 = 4CO2 + 5H2O Now add up the oxygens on the right side. The sum of 13 is odd, so it cannot be balanced as is by the oxygen gas O2. Therefore we must allow in the reaction for 2 molecules of butane and double everything else. 2C4H10 + ?O2 = 8CO2 + 10H2O Now there are 26 atoms of O2, so 13 molecules of O2 gas are reacted: 2C4H10 + 13O2 = 8CO2 + 10H2O
i took help from a site because its not my subject
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it did, thanks :)
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