Use dimensional analysis to show all work: How many grams of excess reagent is used and how many grams remain? A student made 102 g of carbon dioxide what is her % yield? (Rest of work provided on images.)
Use dimensional analysis to show all work: How many grams of excess reagent is used and how many grams remain? A student made 102 g of carbon dioxide what is her % yield? (Rest of work provided on images.)
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question
Use dimensional analysis to show all work:
How many grams of excess reagent is used and how many grams remain?
A student made 102 g of carbon dioxide what is her % yield?
(Rest of work provided on images.)
![### Combustion Reaction of C₆H₁₀
1. **Chemical Equation:**
\[
\text{C}_6\text{H}_{10(g)} + \frac{17}{2} \text{O}_{2(g)} \xrightarrow{\Delta} 6\text{CO}_{2(g)} + 5\text{H}_2\text{O}_{(g)}
\]
### Calculations
2. **For C₆H₁₀:**
- Molar mass of C₆H₁₀:
- \(1 \, \text{mol} = 82 \, \text{g}\)
- Conversion factor: \(\frac{1 \, \text{mol}}{82 \, \text{g}} = 1\)
- \(\frac{82 \, \text{g}}{1 \, \text{mol}} = 1\)
- Converting grams to moles:
\[
45 \, \text{g} \times \frac{1 \, \text{mol}}{82 \, \text{g}} = 0.548 \, \text{mol}
\]
3. **For O₂:**
- Molar mass of O₂:
- \(1 \, \text{mol} = 32 \, \text{g}\)
- Conversion factor: \(\frac{1 \, \text{mol}}{32 \, \text{g}} = 1\)
- \(\frac{32 \, \text{g}}{1 \, \text{mol}} = 1\)
- Converting grams to moles:
\[
156 \, \text{g} \times \frac{1 \, \text{mol}}{32 \, \text{g}} = 4.875 \, \text{mol}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F824a9fa1-de73-4e5c-9ec2-2636af2daa25%2F02865ac0-dd73-4a2b-a34a-c5c996a1db40%2Fhv9m9au_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Combustion Reaction of C₆H₁₀
1. **Chemical Equation:**
\[
\text{C}_6\text{H}_{10(g)} + \frac{17}{2} \text{O}_{2(g)} \xrightarrow{\Delta} 6\text{CO}_{2(g)} + 5\text{H}_2\text{O}_{(g)}
\]
### Calculations
2. **For C₆H₁₀:**
- Molar mass of C₆H₁₀:
- \(1 \, \text{mol} = 82 \, \text{g}\)
- Conversion factor: \(\frac{1 \, \text{mol}}{82 \, \text{g}} = 1\)
- \(\frac{82 \, \text{g}}{1 \, \text{mol}} = 1\)
- Converting grams to moles:
\[
45 \, \text{g} \times \frac{1 \, \text{mol}}{82 \, \text{g}} = 0.548 \, \text{mol}
\]
3. **For O₂:**
- Molar mass of O₂:
- \(1 \, \text{mol} = 32 \, \text{g}\)
- Conversion factor: \(\frac{1 \, \text{mol}}{32 \, \text{g}} = 1\)
- \(\frac{32 \, \text{g}}{1 \, \text{mol}} = 1\)
- Converting grams to moles:
\[
156 \, \text{g} \times \frac{1 \, \text{mol}}{32 \, \text{g}} = 4.875 \, \text{mol}
\]
![### Reaction Calculations: Limiting Reactant Analysis
In this calculation, we’re analyzing the reaction involving C₆H₁₀ and O₂.
#### Step-by-Step Analysis
1. **Initial Reaction Setup**
- From the reaction:
- 1 mol C₆H₁₀ reacts with \(\frac{17}{2}\) mol O₂.
- Therefore, 0.548 mol C₆H₁₀ reacts with:
\[
\frac{17}{2} \times 0.548 \, \text{mol O}_2 = 4.658 \, \text{mol O}_2
\]
2. **Determine the Limiting Reactant**
- The limiting reactant is identified as C₆H₁₀.
3. **Product Calculation**
- 1 mol C₆H₁₀ produces 5 mol CO₂.
- Therefore, 0.548 mol C₆H₁₀ produces:
\[
5 \times 0.548 \, \text{mol CO}_2 = 2.74 \, \text{mol CO}_2
\]
4. **Convert Moles of CO₂ to Grams**
- The molar mass of CO₂ is 44 g/mol.
- Converting 2.74 mol CO₂ to grams:
\[
2.74 \, \text{mol} \times \frac{44 \, \text{g}}{1 \, \text{mol}} = 120.56 \, \text{g}
\]
5. **Conclusion**
- (c) The limiting reactant is C₆H₁₀.
This calculation demonstrates how to identify the limiting reactant and determine the amount of product formed in a chemical reaction.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F824a9fa1-de73-4e5c-9ec2-2636af2daa25%2F02865ac0-dd73-4a2b-a34a-c5c996a1db40%2F4hqgkyh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Reaction Calculations: Limiting Reactant Analysis
In this calculation, we’re analyzing the reaction involving C₆H₁₀ and O₂.
#### Step-by-Step Analysis
1. **Initial Reaction Setup**
- From the reaction:
- 1 mol C₆H₁₀ reacts with \(\frac{17}{2}\) mol O₂.
- Therefore, 0.548 mol C₆H₁₀ reacts with:
\[
\frac{17}{2} \times 0.548 \, \text{mol O}_2 = 4.658 \, \text{mol O}_2
\]
2. **Determine the Limiting Reactant**
- The limiting reactant is identified as C₆H₁₀.
3. **Product Calculation**
- 1 mol C₆H₁₀ produces 5 mol CO₂.
- Therefore, 0.548 mol C₆H₁₀ produces:
\[
5 \times 0.548 \, \text{mol CO}_2 = 2.74 \, \text{mol CO}_2
\]
4. **Convert Moles of CO₂ to Grams**
- The molar mass of CO₂ is 44 g/mol.
- Converting 2.74 mol CO₂ to grams:
\[
2.74 \, \text{mol} \times \frac{44 \, \text{g}}{1 \, \text{mol}} = 120.56 \, \text{g}
\]
5. **Conclusion**
- (c) The limiting reactant is C₆H₁₀.
This calculation demonstrates how to identify the limiting reactant and determine the amount of product formed in a chemical reaction.
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