Calculating the volume of green house emissions...?
The Hazelwood power station in the Latrobe Valley comsumer about 13 million tonnes of coal in one year. The coal used in the power station is composed of approximately 25% carbon. Calculate the volume of the greenhouse gas, carbon dioxide, released each year by the power station (1 mol of carbon dioxide occupies 24.5L at 25 degrees celcius and 1 atm.
Answer:
Taking your Carbon content as is...
25% of 13 = 13/4 = 3.25 (million tons of carbon).
CO2 = 1 mole = 44 ton/mole.
CO2 = 1 mole = 44 ton/mole =24.5L
44 x 24.5 x 3.25 = 3,503 million Litres of CO2. per year.
(Note: Moles can be expressed in ANY MASS UNIT, therefore I've used tons as that is the unit of mass used. However, I'm not assuming that my answer is correct).
There is a major flaw here. A carbon content for coal of 25% is absurdly low; it should be close to 90% (with most of the remainder being calcium and other ash materials). However, for the sake of argument, we'll use the 25% figure. This gives 3.25 million tons of carbon, which will be 3.25 trillion/12 moles of carbon and also of carbon dioxide. Multiply this by 24.5 liters for your answer.
You will also need to know the combustion efficiency of the power plant (heat rate).
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Answer:
Taking your Carbon content as is...
25% of 13 = 13/4 = 3.25 (million tons of carbon).
CO2 = 1 mole = 44 ton/mole.
CO2 = 1 mole = 44 ton/mole =24.5L
44 x 24.5 x 3.25 = 3,503 million Litres of CO2. per year.
(Note: Moles can be expressed in ANY MASS UNIT, therefore I've used tons as that is the unit of mass used. However, I'm not assuming that my answer is correct).
There is a major flaw here. A carbon content for coal of 25% is absurdly low; it should be close to 90% (with most of the remainder being calcium and other ash materials). However, for the sake of argument, we'll use the 25% figure. This gives 3.25 million tons of carbon, which will be 3.25 trillion/12 moles of carbon and also of carbon dioxide. Multiply this by 24.5 liters for your answer.
You will also need to know the combustion efficiency of the power plant (heat rate).
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