Help pleaseeeeeee A 1.946g sample of a component of the light petroleum distilla

Question

Help pleaseeeeeee

A 1.946g sample of a component of the light petroleum distillate called naphtha is found to yield 5.965g CO2(g) and 2.848gH2O(I) on complete combustion. This particular compound is also found to be an alkane with one methyl group attached to a longer carbon chain and to have a molecular formula twice its empirical formula. The compound also has the following properties: melting point of -154 degree C, boiling point of 60.3 degree C. density of 0.6532 g/mL at 20 degree C, specific heat of 2.25 J/(g middle dot C), and

Explanation / Answer

I think you would also need the density of the fuel and its heat of combustion
Heat of combustion is approx 4166 kJ/mol and density = 654.8 kg/m3 (use your own data)
4.18 J/g = 4180 J/kg. Mass of water = density x volume = 654.8 x 22.9 kg

Amount of heat needed = cm[delta]T = 4180 x (29.9 x 654.8) x 13.6
= 1112999705 J
= 1112999.705 kJ

1 mole of hexane weighs approx 86g (use your own data) and releases 4166 kJ of energy.

Mass needed = (86/4166) x 1112999.705 = 22975.99007 g or 22.97599007kg

From density 654.8 kg occupies 1 m3
22.97599007 kg occupies (1/654.8) x 22.97599007 = 0.35088562 m3

This is 350.88562 L which is 350.88562 x (1.057/4) gallons = 9.27215 gallons (I think)

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