Carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) is the gas that is mainly responsible for global warming (the greenhouse effect). The burning of fossil fuels is a major cause of the increased concentration of \(\mathrm{CO}_{2}\) in the atmosphere. Carbon dioxide is also the end product of metabolism (see Example 3.13 ). Using glucose as an example of food, calculate the annual human production of \(\mathrm{CO}_{2}\) in grams, assuming that each person consumes \(5.0 \times 10^{2} \mathrm{~g}\) of glucose per day. The world's population is 7.2 billion, and there are 365 days in a year.

Short Answer

Expert verified
The total annual human production of carbon dioxide is approximately \(1.94 \times 10^{15} \mathrm{g/year}\).

Step by step solution

01

Find the chemical reaction

Consider the chemical reaction where glucose (C6H12O6) is converted into carbon dioxide (CO2) and water (H2O), which is given by \(C6H12O6 + 6O2 -> 6CO2 + 6H2O\). Looking at the stoichiometry, we can see that six moles of carbon dioxide are produced per mol of glucose.
02

Calculate the molar mass of glucose and carbon dioxide

Calculate the molar mass of glucose and carbon dioxide. The molar mass of glucose is \(180.156 \mathrm{g/mol}\) and the molar mass of carbon dioxide is \(44.0095 \mathrm{g/mol}\). Therefore, one gram of glucose produces \((44.0095 \mathrm{g/mol} * 6) / 180.156 = 1.47 \mathrm{g}\) of carbon dioxide.
03

Calculate the daily production of carbon dioxide

To find out the daily carbon dioxide production, multiply the daily consumption of glucose (500 g) by the amount of carbon dioxide produced per gram of glucose. So, each person produces \(500 \mathrm{g} * 1.47 \mathrm{g/g} = 735 \mathrm{g/day}\) of carbon dioxide.
04

Calculate the annual human production of carbon dioxide

Multiply the daily production of carbon dioxide by the world's total population (7.2 billion) and the number of days per year (365) to find out the total annual human production of carbon dioxide: \(735 \mathrm{g/day} * 7.2 \times 10^{9} * 365 = 1.94 \times 10^{15} \mathrm{g/year}\).

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