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# How many moles of water can be produced by burning 325g of octane in excess oxygen?

###### Wiki User

Balanced equation: 2C8H18 + 25O2 ==> 16CO2 + 18H2O
moles of octane used: 325 g x 1 mole/114g = 2.85 moles octane
moles H2O produced: 18 moles H2O/2 moles C8H18 x 2.85 moles C8H18 = 25.65 moles H2O

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###### Wiki User

The chemical reaction is:
2 C8H18 + 25 O2 = 16 CO2 + 18 H2O

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## Related Questions

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OxygenFood (glucose)Starch (excess food/glucose)

Octane + oxygen ---&gt; carbon dioxide + water Lulu

Carbon dioxide and water if excess of oxygen (air) is provided.

Two molecules of octane has 16 carbon atoms. So 16 molecules of carbon dioxide will be formed when two molecules of octane are burned in presence of oxygen.

Heat. This is produced by the chemical reaction between a hydrocarbon and oxygen in the atmosphere.

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Carbon monoxide is produced when a fuel is burning with an insufficient oxygen supply.

It goes into the atmosphere as carbon dioxide. The heat produced by burning fuel comes from forming the carbon-oxygen chemical bonds that creates CO2.

2C8H18 + 25 O2 &rarr; 16CO2 + 18 H2O + energy so the reactants are octane (actually there are other hydrocarbons present too) and oxygen.

The chemical equation for complete burning of octane is: 2 C8H18 + 25 O2 -&gt; 16 CO2 + 18 H2O. This equation shows that 25 moles of diatomic oxygen are required to completely burn each two moles of octane. The gram molecular mass of octane is 114.23 and the gram molecular mass of diatomic oxygen is 2(15.9994). Therefore, the ratio of the mass of oxygen required to completely burn any given mass of octane to the mass of the octane to be burned is 50(15.9994)/2(114.23) or 3.5016, to the justified number of significant digits (the same as the number of digits in the least precisely specified datum 114.23, and the mass of oxygen required to burn 19.8 g of octane is (3.5016)(19.8) or 69.3 grams to the justified number of significant digits, now limited by the less precisely specified datum 119.8.

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