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4. The combustion of methane gas (CH.) is shown in the reaction below: CH4 (g) + O2 (g) → CO2 (g) + H2O (g) If you burn 15 grams of methane with excess oxygen and form 28.4 grams of water, calculate the percent yield of water formed.

4. The combustion of methane gas (CH.) is shown in the reaction below: CH4 (g) + O2 (g) → CO2 (g) + H2O (g) If you burn 15 grams of methane with excess oxygen and form 28.4 grams of water, calculate the percent yield of water formed.

Chemistry
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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### Combustion of Methane Gas (CH₄) The chemical equation for the combustion of methane gas is shown below: \[ \text{CH}_4 (g) + 2\text{O}_2 (g) \rightarrow \text{CO}_2 (g) + 2\text{H}_2\text{O} (g) \] --- **Problem Statement** If you burn 15 grams of methane with excess oxygen and form 28.4 grams of water, calculate the **percent yield of water** formed. --- **Key Concepts** - **Methane Combustion**: Methane reacts with oxygen to produce carbon dioxide and water. - **Stoichiometry**: Utilized to calculate expected product amounts. - **Percent Yield**: It is calculated using the formula: \[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% \] **Calculating Percent Yield** 1. **Determine Molar Masses**: - Methane (CH₄): 16.04 g/mol - Water (H₂O): 18.02 g/mol 2. **Calculate Theoretical Yield**: - Use stoichiometry to find the moles and then mass of water produced. 3. **Use Percent Yield Formula**: - Compare the actual yield (28.4 g) to the theoretical yield to calculate the percent yield. This exercise demonstrates practical application of chemical reactions and stoichiometry in calculating reaction efficiency.
Transcribed Image Text:### Combustion of Methane Gas (CH₄) The chemical equation for the combustion of methane gas is shown below: \[ \text{CH}_4 (g) + 2\text{O}_2 (g) \rightarrow \text{CO}_2 (g) + 2\text{H}_2\text{O} (g) \] --- **Problem Statement** If you burn 15 grams of methane with excess oxygen and form 28.4 grams of water, calculate the **percent yield of water** formed. --- **Key Concepts** - **Methane Combustion**: Methane reacts with oxygen to produce carbon dioxide and water. - **Stoichiometry**: Utilized to calculate expected product amounts. - **Percent Yield**: It is calculated using the formula: \[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% \] **Calculating Percent Yield** 1. **Determine Molar Masses**: - Methane (CH₄): 16.04 g/mol - Water (H₂O): 18.02 g/mol 2. **Calculate Theoretical Yield**: - Use stoichiometry to find the moles and then mass of water produced. 3. **Use Percent Yield Formula**: - Compare the actual yield (28.4 g) to the theoretical yield to calculate the percent yield. This exercise demonstrates practical application of chemical reactions and stoichiometry in calculating reaction efficiency.
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