How many kilograms of Co2 are released during the combustion of 25.00 gallons of

Question

How many kilograms of Co2 are released during the combustion of 25.00 gallons of gasoline? Assume gasoline is pure isoctane with density of 0.9619 g/mL. If the combustion was used to heat 8565 L of water from an initial temperature of 11.0C. What would be the final temperature of the water assuming 54.00% efficiency in the energy transfer?
How many kilograms of Co2 are released during the combustion of 25.00 gallons of gasoline? Assume gasoline is pure isoctane with density of 0.9619 g/mL. If the combustion was used to heat 8565 L of water from an initial temperature of 11.0C. What would be the final temperature of the water assuming 54.00% efficiency in the energy transfer?

Explanation / Answer

1 gallon = 3.785 litres

Thus, 25 gallons = 94.625 litres

Thus, mass of isoctane = density*volume = 0.9619*94625 = 91019.8 g

Moles of gasoline = mass/molar mass = 91019.8/114 = 798.42

Now, C8H18 + (25/2)O2 -----> 8CO2 + 9H2O

Moles of CO2 produced = 8*moles of iso-octane reacting = 8*798.42 = 6387.35

Mass of CO2 produced = moles*molar mass = 6387.35*44 = 281043.5 g = 281.043 kg

Now,

Let the heat of combustion of iso-octane be 'x' J/mole

Thus, total heat produced = 798.42x

Total heat used to heat the water = 0.54*798.42x = 431.15x J

Now, Heat = mass of water*specific heat of water*rise in temperature

or, 431.15x = 8565000*4.184*(T-11)

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