A mixture of butene, C4H8, and butane, C4H10, is burned in air to give CO2 and water. Suppose you burn 2.85 g of the mixture and obtain 8.81 g of CO2 and 3.99 g of H₂O. What are the mass percentages of butene and butane in the mixture? Mass percentage of butene = Mass percentage of butane = % %
A mixture of butene, C4H8, and butane, C4H10, is burned in air to give CO2 and water. Suppose you burn 2.85 g of the mixture and obtain 8.81 g of CO2 and 3.99 g of H₂O. What are the mass percentages of butene and butane in the mixture? Mass percentage of butene = Mass percentage of butane = % %
Chemistry
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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
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![**Combustion Analysis of Butane and Butene Mixture**
In a mixture where butene (\(C_4H_8\)) and butane (\(C_4H_{10}\)) are burned in air, the combustion produces carbon dioxide (\(CO_2\)) and water (\(H_2O\)). For instance, if you burn 2.85 grams of this mixture and obtain 8.81 grams of \(CO_2\) and 3.99 grams of \(H_2O\), you can determine the mass percentages of butene and butane present in the mixture.
To find the mass percentages, we need to use the given information:
1. Total mass of \(CO_2\) produced: 8.81 grams
2. Total mass of \(H_2O\) produced: 3.99 grams
3. Initial mass of the mixture: 2.85 grams
We need to calculate the mass percentages for:
- Butene (\(C_4H_8\))
- Butane (\(C_4H_{10}\))
**Equations:**
1. **Combustion of Butene:**
\[ C_4H_8 + 6O_2 \rightarrow 4CO_2 + 4H_2O \]
2. **Combustion of Butane:**
\[ 2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O \]
Given the stoichiometry of the balanced equations, we first convert the mass of \(CO_2\) and \(H_2O\) produced into moles:
- Molar mass of \(CO_2\): \(44.01 \text{ g/mol}\)
- Molar mass of \(H_2O\): \(18.02 \text{ g/mol}\)
Using these molar masses:
\[ \text{Moles of } CO_2 = \frac{8.81 \text{ g}}{44.01 \text{ g/mol}} \]
\[ \text{Moles of } H_2O = \frac{3.99 \text{ g}}{18.02 \text{ g/mol}} \]
Next, we determine the contributions from butene and butane to the masses of \(CO_2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a354f1c-cdfd-41e5-a863-9a86a504af1c%2Ffc9a13e3-d281-4ff6-b48e-8804a7727f41%2Fysqxk5_processed.png&w=3840&q=75)
Transcribed Image Text:**Combustion Analysis of Butane and Butene Mixture**
In a mixture where butene (\(C_4H_8\)) and butane (\(C_4H_{10}\)) are burned in air, the combustion produces carbon dioxide (\(CO_2\)) and water (\(H_2O\)). For instance, if you burn 2.85 grams of this mixture and obtain 8.81 grams of \(CO_2\) and 3.99 grams of \(H_2O\), you can determine the mass percentages of butene and butane present in the mixture.
To find the mass percentages, we need to use the given information:
1. Total mass of \(CO_2\) produced: 8.81 grams
2. Total mass of \(H_2O\) produced: 3.99 grams
3. Initial mass of the mixture: 2.85 grams
We need to calculate the mass percentages for:
- Butene (\(C_4H_8\))
- Butane (\(C_4H_{10}\))
**Equations:**
1. **Combustion of Butene:**
\[ C_4H_8 + 6O_2 \rightarrow 4CO_2 + 4H_2O \]
2. **Combustion of Butane:**
\[ 2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O \]
Given the stoichiometry of the balanced equations, we first convert the mass of \(CO_2\) and \(H_2O\) produced into moles:
- Molar mass of \(CO_2\): \(44.01 \text{ g/mol}\)
- Molar mass of \(H_2O\): \(18.02 \text{ g/mol}\)
Using these molar masses:
\[ \text{Moles of } CO_2 = \frac{8.81 \text{ g}}{44.01 \text{ g/mol}} \]
\[ \text{Moles of } H_2O = \frac{3.99 \text{ g}}{18.02 \text{ g/mol}} \]
Next, we determine the contributions from butene and butane to the masses of \(CO_2
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