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Determine the mass in grams of CO₂ that are produced by the complete reaction of 11.65 grams of C9H20 (nonane) according to the following combustion reaction: C9H₂0 (1) 14 O₂(g) → 9 CO₂(g) + 10 H₂O(g)

Determine the mass in grams of CO₂ that are produced by the complete reaction of 11.65 grams of C9H20 (nonane) according to the following combustion reaction: C9H₂0 (1) 14 O₂(g) → 9 CO₂(g) + 10 H₂O(g)

Chemistry
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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**Question 12 of 18** **Determine the mass in grams of CO₂ that are produced by the complete reaction of 11.65 grams of C₉H₂₀ (nonane) according to the following combustion reaction:** C₉H₂₀(l) + 14 O₂(g) → 9 CO₂(g) + 10 H₂O(g) --- **Explanation:** This problem involves a stoichiometric calculation for the combustion of nonane (C₉H₂₀). In this exercise, the goal is to determine how many grams of carbon dioxide (CO₂) are formed when 11.65 grams of nonane undergo complete combustion. **Key Components:** 1. **Starting Amount:** You begin with 11.65 grams of C₉H₂₀. 2. **Combustion Reaction:** The balanced chemical equation for the combustion reaction shows 1 mole of nonane reacts with 14 moles of oxygen to produce 9 moles of carbon dioxide and 10 moles of water. 3. **Calculation Grid:** - Various values are provided for use in stoichiometric calculations, including molecular weights and mole coefficients: - Molecular weight of CO₂: 44.01 g/mol - Molecular weight of C₉H₂₀: 128.2 g/mol - Mole ratio for CO₂: 9 - Mole ratio for O₂: 14 - Mole ratio for H₂O: 10 - Avogadro's number: \(6.022 \times 10^{23}\) **Process:** To find the grams of CO₂ produced: - Convert grams of C₉H₂₀ to moles using its molecular weight. - Use the mole ratio from the balanced equation to find moles of CO₂ produced. - Convert moles of CO₂ to grams using its molecular weight. **Additional Note:** The grid also includes terms and values like: - “Add Factor”: To include additional calculations if needed. - “Answer” and “Reset”: Sections to input and check answers, or reset the calculations. This explanation helps guide a student or learner through the process of solving a stoichiometry problem involving the combustion of hydrocarbons.
Transcribed Image Text:**Question 12 of 18** **Determine the mass in grams of CO₂ that are produced by the complete reaction of 11.65 grams of C₉H₂₀ (nonane) according to the following combustion reaction:** C₉H₂₀(l) + 14 O₂(g) → 9 CO₂(g) + 10 H₂O(g) --- **Explanation:** This problem involves a stoichiometric calculation for the combustion of nonane (C₉H₂₀). In this exercise, the goal is to determine how many grams of carbon dioxide (CO₂) are formed when 11.65 grams of nonane undergo complete combustion. **Key Components:** 1. **Starting Amount:** You begin with 11.65 grams of C₉H₂₀. 2. **Combustion Reaction:** The balanced chemical equation for the combustion reaction shows 1 mole of nonane reacts with 14 moles of oxygen to produce 9 moles of carbon dioxide and 10 moles of water. 3. **Calculation Grid:** - Various values are provided for use in stoichiometric calculations, including molecular weights and mole coefficients: - Molecular weight of CO₂: 44.01 g/mol - Molecular weight of C₉H₂₀: 128.2 g/mol - Mole ratio for CO₂: 9 - Mole ratio for O₂: 14 - Mole ratio for H₂O: 10 - Avogadro's number: \(6.022 \times 10^{23}\) **Process:** To find the grams of CO₂ produced: - Convert grams of C₉H₂₀ to moles using its molecular weight. - Use the mole ratio from the balanced equation to find moles of CO₂ produced. - Convert moles of CO₂ to grams using its molecular weight. **Additional Note:** The grid also includes terms and values like: - “Add Factor”: To include additional calculations if needed. - “Answer” and “Reset”: Sections to input and check answers, or reset the calculations. This explanation helps guide a student or learner through the process of solving a stoichiometry problem involving the combustion of hydrocarbons.
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