Problem 31
Question
Land Management It has been estimated that better management of cropland, grazing land, and forests could reduce the amount of carbon dioxide in the atmosphere by \(5.4 \times 10^{9}\) kilograms of carbon per year. a. How many moles of carbon are present in \(5.4 \times 10^{9}\) kilograms of carbon? b. How many kilograms of carbon dioxide does this quantity of carbon represent?
Step-by-Step Solution
Verified Answer
Answer: 1.98 x 10^10 kg of carbon dioxide
1Step 1: Convert the mass of carbon into moles
First, we need to convert the mass of carbon into moles. To do this, we will use the molar mass of carbon which is 12 g/mol. The given mass of carbon is \(5.4 \times 10^{9}\) kg, which is equal to \(5.4 \times 10^{12}\) grams.
The formula to convert the mass of carbon into moles is:
moles = mass / molar mass
moles = \(\frac{5.4 \times 10^{12} grams}{12 grams/mol}\)
2Step 2: Calculate the moles of carbon
Now, we just need to perform the calculation:
moles = \(\frac{5.4 \times 10^{12} grams}{12 grams/mol}\) = \(4.5 \times 10^{11}\) moles
So, there are \(4.5 \times 10^{11}\) moles of carbon present in \(5.4 \times 10^{9}\) kilograms of carbon.
3Step 3: Convert the moles of carbon to moles of carbon dioxide
Since one mole of carbon combines with one mole of oxygen to form one mole of carbon dioxide, the moles of carbon dioxide will be equal to the moles of carbon.
moles of carbon dioxide = moles of carbon = \(4.5 \times 10^{11}\) moles
4Step 4: Calculate the mass of carbon dioxide
To find the mass of carbon dioxide, we will use the molar mass of carbon dioxide, which is 44 g/mol. We will multiply the moles of carbon dioxide by its molar mass:
mass of carbon dioxide = moles of carbon dioxide × molar mass of carbon dioxide
mass of carbon dioxide = \(4.5 \times 10^{11}\) moles × 44 g/mol
5Step 5: Convert the mass of carbon dioxide into kilograms
Finally, let's perform the calculation and convert the mass into kilograms:
mass of carbon dioxide = \(4.5 \times 10^{11}\) moles × 44 g/mol = \(1.98 \times 10^{13}\) grams
mass of carbon dioxide (in kilograms) = \(\frac{1.98 \times 10^{13} grams}{1000}\) = \(1.98 \times 10^{10}\) kg
So, the given quantity of carbon (\(5.4 \times 10^{9}\) kg) represents \(1.98 \times 10^{10}\) kg of carbon dioxide.
Key Concepts
Mole ConversionCarbon DioxideMolar MassCarbon Emissions
Mole Conversion
Mole conversion is a crucial concept in stoichiometry, helping us to connect mass with the number of particles. When we talk about moles, we're referring to the number of atoms or molecules in a substance. One mole equals Avogadro’s number, which is approximately \(6.022 \times 10^{23}\) particles. This relationship allows chemists to calculate the amount of substance present based on its mass.
To convert grams into moles, we use the formula:
To convert grams into moles, we use the formula:
- moles = mass (in grams) / molar mass (g/mol)
Carbon Dioxide
Carbon dioxide (CO₂) is a compound composed of one carbon atom and two oxygen atoms. It is a major greenhouse gas contributing to global warming through the greenhouse effect. Understanding how carbon turns into carbon dioxide is vital for environmental and chemistry studies.
In the exercise, one mole of carbon combined with one mole of O₂ forms one mole of CO₂. This conversion reflects the stoichiometry of their reaction, which is represented as:
In the exercise, one mole of carbon combined with one mole of O₂ forms one mole of CO₂. This conversion reflects the stoichiometry of their reaction, which is represented as:
- \(\text{C} + \text{O}_2 \rightarrow \text{CO}_2\)
Molar Mass
Molar mass is the mass of one mole of a given substance, generally expressed in grams per mole (g/mol). It is a bridge between how much a substance weighs and the amount of its particles. Molar mass allows us to translate between the macroscopic world we can measure and the microscopic world of atoms and molecules.
For the problem provided, knowing the molar mass of carbon (12 g/mol) and carbon dioxide (44 g/mol) is essential:
For the problem provided, knowing the molar mass of carbon (12 g/mol) and carbon dioxide (44 g/mol) is essential:
- Molar mass of carbon: 12 g/mol
- Molar mass of carbon dioxide: 44 g/mol (12 from carbon + 16×2 from oxygen)
Carbon Emissions
Carbon emissions refer to the release of carbon, particularly as carbon dioxide, into the atmosphere from various activities. They are significant contributors to global warming and climate change. Reducing carbon emissions is a critical environmental goal that can be achieved through better land management, as hinted in the exercise.
By converting the mass of carbon to carbon dioxide, we can visualize the impact of carbon emissions. In this exercise, understanding how much CO₂ a certain amount of carbon represents is key to assessing and controlling emissions. Quantifying these emissions in kilograms emphasizes the scale of the issue. An accurate measure helps in developing strategies to mitigate atmospheric CO₂ levels, reducing the adverse effects on the climate.
By converting the mass of carbon to carbon dioxide, we can visualize the impact of carbon emissions. In this exercise, understanding how much CO₂ a certain amount of carbon represents is key to assessing and controlling emissions. Quantifying these emissions in kilograms emphasizes the scale of the issue. An accurate measure helps in developing strategies to mitigate atmospheric CO₂ levels, reducing the adverse effects on the climate.
Other exercises in this chapter
Problem 29
There are two ways to write the equation for the combustion of ethane: $$\begin{aligned}\mathrm{C}_{2} \mathrm{H}_{6}(g)+\frac{7}{2} \mathrm{O}_{2}(g) \rightarr
View solution Problem 30
Suppose the same mass of each of these components of natural gas was completely combusted: (a) \(\mathrm{CH}_{4},\) (b) \(\mathrm{C}_{2} \mathrm{H}_{6}\) (c) \(
View solution Problem 32
Most of the \(\mathrm{CO}_{2}\) emitted by industrial sources comes from the combustion of fossil fuels, but some is also produced by cement manufacturing and t
View solution Problem 33
When \(\mathrm{NaHCO}_{3}\) is heated above \(270^{\circ} \mathrm{C},\) it decomposes to \(\mathrm{Na}_{2} \mathrm{CO}_{3}(s), \mathrm{H}_{2} \mathrm{O}(g),\) a
View solution