How many grams of octane (C8H18) must be burned to produce 300.0g of CO2?
1 answer:
Answer:
Explanation:
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In this case, according to the given combustion reaction of octane, it is possible for us to perform the stoichiometric method in order to calculate the mass of octane that is required to consume 300.0 g of oxygen by considering the 2:25 mole ratio, and the molar masses of 114.22 g/mol and 32.00 g/mol respectively:
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Answer:
Explanation:
You will need to use the conversion between food calories, calories and joules.
From a nutritional table (from the internet) you can obtain the amount of nutritional calories for each type macronutrient:<em> carbohydrates, protein, and fat.</em>
Macronutrient food calories
Carbohydrate 4 cal/g
Protein 4 cal/g
Fat 9 cal/g
To convert food calories to regular calories, mutiply by the conversion factor:
Macronutrient calories
Carbohydrate 4,000 cal/g
Protein 4,000 cal/g
Fat 9,000 cal/g
To convert regular calories to joules, multiply by the conversion factor:
Macronutrient energy in joules
Carbohydrate 16,736 joule/g
Protein 16,736 joule/g
Fat 37,656 joule/g
Now you can multiply the grams of each macronutrient in one potato by its energy content:
Macronutrient energy/gram × grams = total energy
Carbohydrate 16,736 joule/g × 37 g = 619,232 joules
Protein 16,736 joule/g × 4.3 g = 71,968.8 joules
Fat 37,656 joule/g × 4 g = 150,624 joules
Total = 619,232 joules + 71,968.8 joules + 150,624 joules = 841,820.8 joules.
Since they are <u>two potatoes</u>, multiply by 2: × 841,820.8 joules = 1,683,641.6 joules.
Divide by 1,000 to convert to kilojoules = 1,683.64 kilojoules
Divide the total energy from the two potatoes by the <em>energy that an average person expend per hour of walking (845 kilojoules of energy per hour):</em>