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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 Sample Problem 3.4). 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, that the world's population is 6.5 billion, and that there are 365 days in a year.

Short Answer

Expert verified
The total annual human production of CO2 is approximately 1.57 × 10^14 grams.

Step by step solution

01

Write the Balanced Chemical Equation for Glucose Metabolism

The chemical reaction for the metabolism of glucose can be represented as:\[\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \rightarrow 6\text{CO}_2 + 6\text{H}_2\text{O}\] This equation tells us that one mole of glucose metabolizes to produce six moles of carbon dioxide.
02

Determine Moles of Glucose Consumed per Person Annually

First, calculate the moles of glucose consumed daily by a person:- Molar mass of glucose \(\text{C}_6\text{H}_{12}\text{O}_6\) is \(180 \text{ g/mol}\).- Glucose consumed per day: \(5.0 \times 10^2 \text{ g}\).Moles of glucose per day = \(\frac{5.0 \times 10^2}{180}\).Moles of glucose annually per person = \(\left(\frac{5.0 \times 10^2}{180}\right) \times 365\).
03

Calculate Annual Production of CO2 Per Person

From the balanced equation, 1 mole of glucose produces 6 moles of \(\text{CO}_2\).Therefore, CO2 moles per person per year = \(6 \times \left(\frac{5.0 \times 10^2}{180}\right) \times 365\).
04

Convert Moles of CO2 to Grams per Person

The molar mass of \(\text{CO}_2\) is \(44 \text{ g/mol}\).Grams of \(\text{CO}_2\) per person per year = \(44 \times 6 \times \left(\frac{5.0 \times 10^2}{180}\right) \times 365\).
05

Calculate Total Global Annual Production of CO2

Given the world population as 6.5 billion, the total CO2 production is:Total grams of \(\text{CO}_2\) per year = \(6.5 \times 10^9 \times 44 \times 6 \times \left(\frac{5.0 \times 10^2}{180}\right) \times 365\).This simplifies to calculate the total CO2 production.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Glucose Metabolism
Glucose metabolism is a crucial process in the human body that transforms food into energy. It involves a sequence where glucose (sugar) is broken down to produce energy for bodily functions. The chemical equation that represents this process is:\[\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \rightarrow 6\text{CO}_2 + 6\text{H}_2\text{O}\]In this reaction, one molecule of glucose reacts with oxygen
  • This produces carbon dioxide (\(\text{CO}_2\)) and water (\(\text{H}_2\text{O}\)).
  • The process releases energy needed for cellular activities.
Understanding glucose metabolism is key in assessing how nutrients like glucose are efficiently used by our bodies and the balance of gases produced in these metabolic reactions.
Balanced Chemical Equation
A balanced chemical equation is a vital concept in chemistry. It ensures the number of atoms for each element is conserved throughout the reaction. In the context of glucose metabolism:\[\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \rightarrow 6\text{CO}_2 + 6\text{H}_2\text{O}\]This shows:
  • One glucose molecule produces six molecules of carbon dioxide.
  • Six oxygen molecules are needed to metabolize one glucose molecule.
This balance is crucial for determining the ratio of reactants and products and is helpful in calculating quantities like the moles of carbon dioxide produced in metabolism.
Moles of Glucose
Understanding moles of glucose is important to calculating chemical reactions. The concept of a mole helps quantify substances:- Molar mass of glucose (\(\text{C}_6\text{H}_{12}\text{O}_6\)) is \(180 \text{ g/mol}\).- Daily glucose consumption: \(5.0 \times 10^2 \text{ g}\).To find the moles of glucose consumed daily, you use this formula:\[\text{Moles of glucose per day} = \frac{5.0 \times 10^2}{180}\]This indicates how much glucose individuals metabolize and sets the foundation for further calculations of products like carbon dioxide in metabolism.
Molar Mass Calculation
Molar mass calculation is a basic yet crucial tool in chemistry to convert between grams and moles. It allows determining the mass of each element in a compound. For example:- Glucose (\(\text{C}_6\text{H}_{12}\text{O}_6\)) has a molar mass of \(180 \text{ g/mol}\).- Carbon dioxide (\(\text{CO}_2\)) has a molar mass of \(44 \text{ g/mol}\).When many moles of a compound like glucose or carbon dioxide are involved, knowing the molar mass:
  • Helps convert moles into grams.
  • Facilitates precise calculations in reactions, such as knowing how much \(\text{CO}_2\) is produced from metabolizing glucose.
Essentially, mastering molar mass calculations enables accurate predictions in chemistry regarding the amount of products formed from given reactants.

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Most popular questions from this chapter

When potassium cyanide ( \(\mathrm{KCN}\) ) reacts with acids, a deadly poisonous gas, hydrogen cyanide (HCN), is given off. Here is the equation: $$ \mathrm{KCN}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{KCl}(a q)+\mathrm{HCN}(g) $$ If a sample of \(0.140 \mathrm{~g}\) of \(\mathrm{KCN}\) is treated with an excess of \(\mathrm{HCl}\), calculate the amount of HCN formed, in grams.

A certain sample of coal contains 1.6 percent sulfur by mass. When the coal is burned, the sulfur is converted to sulfur dioxide. To prevent air pollution, this sulfur dioxide is treated with calcium oxide \((\mathrm{CaO})\) to form calcium sulfite \(\left(\mathrm{CaSO}_{3}\right) .\) Calculate the daily mass (in kilograms) of \(\mathrm{CaO}\) needed by a power plant that uses \(6.60 \times 10^{6} \mathrm{~kg}\) of coal per day.

When heated, lithium reacts with nitrogen to form lithium nitride: $$ 6 \mathrm{Li}(s)+\mathrm{N}_{2}(g) \stackrel{\Delta}{\longrightarrow} 2 \mathrm{Li}_{3} \mathrm{~N}(s) $$ What is the theoretical yield of \(\mathrm{Li}_{3} \mathrm{~N}\) in grams when \(12.3 \mathrm{~g}\) of \(\mathrm{Li}\) is heated with \(33.6 \mathrm{~g}\) of \(\mathrm{N}_{2}\) ? If the actual yield of \(\mathrm{Li}_{2} \mathrm{~N}\) is \(5.89 \mathrm{~g}\), what is the percent vield of the reaction?

Determine whether each of the following equations represents a combination reaction, a decomposition reaction, or a combustion reaction: (a) \(2 \mathrm{NaHCO}_{3} \longrightarrow\) \(\mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O},(\mathrm{b}) \mathrm{NH}_{3}+\mathrm{HCl} \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}\) (c) \(2 \mathrm{CH}_{3} \mathrm{OH}+3 \mathrm{O}_{2} \longrightarrow 2 \mathrm{CO}_{2}+4 \mathrm{H}_{2} \mathrm{O}\)

Calculate the percent composition by mass of all the elements in calcium phosphate \(\left[\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\right]\), a major component of bone.

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