Can someone please solve this Organic Chemistry Calculation.
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Can someone please solve this Organic Chemistry Calculation.

[From: ] [author: ] [Date: 11-10-22] [Hit: ]
0*0.703 = 42.Molar mass C8H18 = 114.114.2 g C8H18 will produce 352g CO2 OR Mass ratio = 114.42.......
Petrol from vehicles contains hydrocarbons with from five to twelve carbon atoms. If we assume the average number of carbon atoms per molecule is 8, then petrol can be represented as octane. Using this assumption, calculate the mass of carbon dioxide released into the atmosphere if all the petrol in the 60.0L fuel tank is burnt in its engine. (1.00L of octane has a mass of 0.703kg)

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Balanced equation:
2 C8H18 + 25 O2 = 16 CO2 + 18 H2O
1mol C8H18 will produce 8 mol CO2

Mass of 60.0L octane = 60.0*0.703 = 42.18kg
Molar mass C8H18 = 114.23 g/mol
Molar mass CO2 = 44g/mol : 8 mol = 8*44 = 352g

114.2 g C8H18 will produce 352g CO2 OR Mass ratio = 114.2:352
42.18kg C8H18 will produce 42.18*352/114.2 = 130.0kg CO2 produced.

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Octane is C8H18

C8H18 + 17.5 O2 ---> 8CO2 + 9H2O

The MW of C8H18 is 114g/mole (12*8 + 18*1)

The liter weighs 703g.

703g * 1 mole/114g * 8 moles CO2/1 moles C8H18 = 6.17 moles CO2

I assume they want the volume at STP. 22.4 liter/mole.

6.17 moles CO2 * 22.4 liter/mole = 138 liters CO2 @ STP

That's for one liter of Octane. Forgot to multiply by 60. It's a 60 liter tank.

138 liters CO2/1 liter C8H18 * 60 liters C8H18 = 8,280 liters CO2
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