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researcher88

help needed

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I have tried to do this question serveral times but i dont seem to get the right answer...any help will be appreciated.

...BTW my chemistry teacher thinks its a hard question.

 

If petrol sells 80 cents per litre, what would the price of LPG needed to be for it to have an equivalent price in cents per kilojoule as petrol?

 

petrol heat of combustion:5460 kj/mol

LPG heat of combustion 2220 kj/mol

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To solve this you need to know the density of petrol and also the average molecular weight of petrol and LPG.

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Just know the molar mass of LPG and petrol

Without density you can't do anything !

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The density of petrol is 0.70g/ml and LPG is 0.49g/ml

What about average molar mass ?

You'd need that !!

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ya, you are right.In my mind, i think molar mass and molar volume

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Can't work numericals in mind, will give you exact answer when I get home and get hold of a calculator.

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petrol heat of combustion:5460 kj/mol

LPG heat of combustion 2220 kj/mol

 

 

2 C8H18 + 25 O2 = 18 H2O + 16 CO2 (combustion of octane)

 

CH3-CH2-CH3 + 5 O2 = 3 CO2 + 4 H2O (combustion of propane)

 

1 mole of propane combusts

 

2 mole of Octane combusts

 

Octane uses 10920 Kj

 

Propane uses 2220 Kj

 

So if Octane is 80 cents per 10920Kj

 

Octane = 80 / 10920 = 0.00733 cents per Kj

 

You get 2 moles of propane for the equivalent of octane

 

So 0.00733 / 2 = 0.0037

 

So 0.0037 cents per kj

 

And 2220 kj = 8.2 cents

 

I am not sure if this is correct, it’s early in the morning

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If x is the price in cents per litre, d the density in Kg/lt, H the heat of combustion in KJ/mol and M the molar mass in gms..........

[MATH]\frac{H_p*x_p*M_p}{d_p}=\frac{H_n*x_n*M_n}{d_n}[/MATH]

The subscript p for petrol and n for LPG.

( I basically equated the price in cents/KJ)

Plug in ther constants and given quntities to get your answer........

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