

Subject:
Combustion of Octane
Category: Science Asked by: teriberryga 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 +25O216CO2+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) 

Subject:
Re: Combustion of Octane
Answered By: liviofloresga on 12 Nov 2006 13:36 PST 
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, liviofloresga  


Subject:
Re: Combustion of Octane
From: momomcga on 24 Nov 2006 10:03 PST 
The equation is not balanced. Calculations must be based on a balanced equation to be correct. 
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