Predict how many moles of CaO he should have ended up with, using the following steps. Given that your friend started with 5.0 g of CaCO3 (molar mass = 100 g/mol), Calculate the number of moles in his 5.0 g starting CaCO3 sample, then use your answer to convert from moles of CaCO3 to moles of CaO he should have produced.
Predict how many moles of CaO he should have ended up with, using the following steps. Given that your friend started with 5.0 g of CaCO3 (molar mass = 100 g/mol), Calculate the number of moles in his 5.0 g starting CaCO3 sample, then use your answer to convert from moles of CaCO3 to moles of CaO he should have produced.
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Predict how many moles of CaO he should have ended up with, using the following steps. Given that your friend started with 5.0 g of CaCO3 (molar mass = 100 g/mol), Calculate the number of moles in his 5.0 g starting CaCO3 sample, then use your answer to convert from moles of CaCO3 to moles of CaO he should have produced.
![**Understanding Chemical Reactions: The Decomposition of Calcium Carbonate (CaCO3)**
The molecule CaCO₃ is what makes up eggshells, and it can decompose (break down) when heated strongly. This decomposition reaction is given by the equation:
\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \]
Your friend argues that the chemical equation doesn't follow the law of conservation of mass. He says, "Moles are not conserved since 1 mole of reactants produced 2 moles of product. This means the mass is higher in the end." You think he’s wrong, and want to explain and do an experiment to prove it.
---
**Explanation:**
When we look at the chemical equation, it shows that one mole of calcium carbonate (\( \text{CaCO}_3 \)) decomposes into one mole of calcium oxide (\( \text{CaO} \)) and one mole of carbon dioxide (\( \text{CO}_2 \)). It may seem initially that there are more moles of products (2 moles) compared to the reactants (1 mole). However, it is essential to understand that the law of conservation of mass states that mass must remain constant in a closed system during a chemical reaction. The number of moles may differ, but the mass of the reactants and products must balance.
In this decomposition reaction:
- \( \text{CaCO}_3 \) breaks down into both \( \text{CaO} \) and \( \text{CO}_2 \).
- Instead of counting moles, consider the molar mass: \( \text{CaCO}_3 \) (calcium carbonate) has a molar mass equal to the sum of the molar masses of \( \text{CaO} \) and \( \text{CO}_2 \).
**To Demonstrate and Experiment:**
To prove the conservation of mass, perform an experiment where you measure the mass of \( \text{CaCO}_3 \) before heating. After decomposition, measure the mass of \( \text{CaO} \) left in the container and the \( \text{CO}_2 \) generated (which can be captured in a separate apparatus). The sum of the masses of \( \text{CaO} \) and \( \text{CO}_](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F46ade868-a4ca-4dcd-b69f-bba1f121fc8f%2Ffe698020-aeb0-450d-bf29-c98fe49286d1%2F6cgxcj7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Understanding Chemical Reactions: The Decomposition of Calcium Carbonate (CaCO3)**
The molecule CaCO₃ is what makes up eggshells, and it can decompose (break down) when heated strongly. This decomposition reaction is given by the equation:
\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \]
Your friend argues that the chemical equation doesn't follow the law of conservation of mass. He says, "Moles are not conserved since 1 mole of reactants produced 2 moles of product. This means the mass is higher in the end." You think he’s wrong, and want to explain and do an experiment to prove it.
---
**Explanation:**
When we look at the chemical equation, it shows that one mole of calcium carbonate (\( \text{CaCO}_3 \)) decomposes into one mole of calcium oxide (\( \text{CaO} \)) and one mole of carbon dioxide (\( \text{CO}_2 \)). It may seem initially that there are more moles of products (2 moles) compared to the reactants (1 mole). However, it is essential to understand that the law of conservation of mass states that mass must remain constant in a closed system during a chemical reaction. The number of moles may differ, but the mass of the reactants and products must balance.
In this decomposition reaction:
- \( \text{CaCO}_3 \) breaks down into both \( \text{CaO} \) and \( \text{CO}_2 \).
- Instead of counting moles, consider the molar mass: \( \text{CaCO}_3 \) (calcium carbonate) has a molar mass equal to the sum of the molar masses of \( \text{CaO} \) and \( \text{CO}_2 \).
**To Demonstrate and Experiment:**
To prove the conservation of mass, perform an experiment where you measure the mass of \( \text{CaCO}_3 \) before heating. After decomposition, measure the mass of \( \text{CaO} \) left in the container and the \( \text{CO}_2 \) generated (which can be captured in a separate apparatus). The sum of the masses of \( \text{CaO} \) and \( \text{CO}_
![### Bonus Exercise: Mole Calculation of CaO
**Problem Statement:**
Predict how many moles of CaO (Calcium Oxide) your friend should have ended up with, using the following steps. Given that your friend started with 5.0 g of CaCO₃ (Calcium Carbonate) with a molar mass of 100 g/mol, follow these steps:
1. **Calculate the number of moles in his 5.0 g starting CaCO₃ sample.**
2. **Use your answer to convert from moles of CaCO₃ to moles of CaO he should have produced.**
**Steps for Calculation:**
1. **Calculate the number of moles in the 5.0 g CaCO₃ sample.**
To find the number of moles, use the formula:
\[
\text{Number of moles} = \frac{\text{Mass of sample (g)}}{\text{Molar mass (g/mol)}}
\]
Substituting the given values:
\[
\text{Number of moles of CaCO₃} = \frac{5.0 \, \text{g}}{100 \, \text{g/mol}} = 0.05 \, \text{moles}
\]
2. **Convert from moles of CaCO₃ to moles of CaO.**
The decomposition of CaCO₃ yields CaO and CO₂ in a 1:1 molar ratio, as shown by the balanced chemical equation:
\[
\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2
\]
Therefore:
\[
\text{Moles of CaO} = \text{Moles of CaCO}_3
\]
Given that the moles of CaCO₃ are 0.05 moles:
\[
\text{Moles of CaO} = 0.05 \, \text{moles}
\]
**Conclusion:**
From 5.0 g of CaCO₃, your friend should have produced 0.05 moles of CaO.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F46ade868-a4ca-4dcd-b69f-bba1f121fc8f%2Ffe698020-aeb0-450d-bf29-c98fe49286d1%2F1a1vm3h_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Bonus Exercise: Mole Calculation of CaO
**Problem Statement:**
Predict how many moles of CaO (Calcium Oxide) your friend should have ended up with, using the following steps. Given that your friend started with 5.0 g of CaCO₃ (Calcium Carbonate) with a molar mass of 100 g/mol, follow these steps:
1. **Calculate the number of moles in his 5.0 g starting CaCO₃ sample.**
2. **Use your answer to convert from moles of CaCO₃ to moles of CaO he should have produced.**
**Steps for Calculation:**
1. **Calculate the number of moles in the 5.0 g CaCO₃ sample.**
To find the number of moles, use the formula:
\[
\text{Number of moles} = \frac{\text{Mass of sample (g)}}{\text{Molar mass (g/mol)}}
\]
Substituting the given values:
\[
\text{Number of moles of CaCO₃} = \frac{5.0 \, \text{g}}{100 \, \text{g/mol}} = 0.05 \, \text{moles}
\]
2. **Convert from moles of CaCO₃ to moles of CaO.**
The decomposition of CaCO₃ yields CaO and CO₂ in a 1:1 molar ratio, as shown by the balanced chemical equation:
\[
\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2
\]
Therefore:
\[
\text{Moles of CaO} = \text{Moles of CaCO}_3
\]
Given that the moles of CaCO₃ are 0.05 moles:
\[
\text{Moles of CaO} = 0.05 \, \text{moles}
\]
**Conclusion:**
From 5.0 g of CaCO₃, your friend should have produced 0.05 moles of CaO.
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