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Q: Combustion of Octane ( Answered,   1 Comment ) Question
 Subject: Combustion of Octane Category: Science Asked by: teriberry-ga List Price: \$2.00 Posted: 12 Nov 2006 12:58 PST Expires: 12 Dec 2006 12:58 PST Question ID: 782154
 ```The equation for the complete combustion of octane is as follows 2C8H18 +25O2-----16CO2+H20 When 228g of octane are burnt completey. Calculate 1. the volume of oxygen needed 2. the volume of carbon dioxide produced (Assume that 1 mole of gas has a volume of 22.4dm3)``` ```Hi!! The equation gives the moles ratio between reactants and products. For case, to burn 2 moles of C8H18, 25 moles of O2 are needed, 16 moles of CO2 and 1 single mole of water are produced. 1 mole of octane weights (8*12g + 18*1g) = 114g So 228g of octane are equal to (228/114) = 2 moles of octane; then if the 228g of octane are burnt completely; this means that 2 moles of octane are burnt completely; therefore 25 moles of O2 are needed. Since each mole has a volume of 22.4dm3, the volume of oxygen needed is (25*22.4dm3) = 560dm3 As a product of the reaction 16 moles of CO2 are produced, this amount of CO2 has a volume of (16*22.4dm3) = 358.4dm3 I hope this helps you. Regards, livioflores-ga``` Clarification of Answer by livioflores-ga on 24 Nov 2006 17:55 PST ```Hi!! The commenter momomc-ga is right, the balanced equation is: 2C8H18 + 25O2 ---> 16CO2 + 18H2O Affortunately the only part that is wrong in the original misbalanced equation was the "+ 18H2O", specifically the number of moles of water that the combustion produces. Since the other terms of the equation are right and no question was made about the water, the answer does not change. Regards, livioflores-ga``` ```The equation is not balanced. Calculations must be based on a balanced equation to be correct.``` 