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Part H Calculate the grams of H20 that would be produced. Express your answer with the appropriate units. HA ? m(H2O) = Value Units

Part H Calculate the grams of H20 that would be produced. Express your answer with the appropriate units. HA ? m(H2O) = Value Units

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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**Part H** Calculate the grams of H\(_2\)O that would be produced. *Express your answer with the appropriate units.* \[ \text{m(H}_2\text{O)} = \] - **Value**: [Input box] - **Units**: [Input box] [Submit Button] [Previous Answers] [Request Answer] --- **Attempt Feedback** **Incorrect; Try Again; 5 attempts remaining** Start by determining the limiting reagent (Part G), and then use the mass of this reagent to find the mass of product. When given a mass of reagent and asked to find the mass of product formed, start by converting the mass to moles using the molar mass of the reagent in g/mol. Once you have the number of moles of reagent, use the balanced chemical equation to find the number of moles of product formed. Finally, convert the moles of product formed to a mass using its molar mass. [Next Button] --- **Note:** This section is part of a chemistry exercise focused on stoichiometry skills, specifically calculating mass from chemical reactions using limiting reagents.
Transcribed Image Text:**Part H** Calculate the grams of H\(_2\)O that would be produced. *Express your answer with the appropriate units.* \[ \text{m(H}_2\text{O)} = \] - **Value**: [Input box] - **Units**: [Input box] [Submit Button] [Previous Answers] [Request Answer] --- **Attempt Feedback** **Incorrect; Try Again; 5 attempts remaining** Start by determining the limiting reagent (Part G), and then use the mass of this reagent to find the mass of product. When given a mass of reagent and asked to find the mass of product formed, start by converting the mass to moles using the molar mass of the reagent in g/mol. Once you have the number of moles of reagent, use the balanced chemical equation to find the number of moles of product formed. Finally, convert the moles of product formed to a mass using its molar mass. [Next Button] --- **Note:** This section is part of a chemistry exercise focused on stoichiometry skills, specifically calculating mass from chemical reactions using limiting reagents.
**Part G: Limiting Reactant Determination** **Problem Statement:** Determine the limiting reactant, given 31.0 g of each reactant in the following chemical equation: \[ \text{C}_2\text{H}_6\text{O (l)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 3\text{H}_2\text{O (g)} \] **Options:** - H₂O - C₂H₆O - O₂ - CO₂ **Solution:** The correct answer involves calculating the number of moles of CO₂ produced from each reactant. 1. **Calculation using C₂H₆O:** - Moles of CO₂ formed \( = 31.0 \, \text{g C}_2\text{H}_6\text{O} \times \frac{1 \, \text{mol C}_2\text{H}_6\text{O}}{46.07 \, \text{g}} \times \frac{2 \, \text{mol CO}_2}{1 \, \text{mol C}_2\text{H}_6\text{O}} = 1.35 \, \text{mol CO}_2 \) 2. **Calculation using O₂:** - Moles of CO₂ formed \( = 31.0 \, \text{g O}_2 \times \frac{1 \, \text{mol O}_2}{32.00 \, \text{g}} \times \frac{2 \, \text{mol CO}_2}{3 \, \text{mol O}_2} = 0.646 \, \text{mol CO}_2 \) **Conclusion:** Since the number of moles produced by O₂ is less, O₂ is the limiting reagent.
Transcribed Image Text:**Part G: Limiting Reactant Determination** **Problem Statement:** Determine the limiting reactant, given 31.0 g of each reactant in the following chemical equation: \[ \text{C}_2\text{H}_6\text{O (l)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 3\text{H}_2\text{O (g)} \] **Options:** - H₂O - C₂H₆O - O₂ - CO₂ **Solution:** The correct answer involves calculating the number of moles of CO₂ produced from each reactant. 1. **Calculation using C₂H₆O:** - Moles of CO₂ formed \( = 31.0 \, \text{g C}_2\text{H}_6\text{O} \times \frac{1 \, \text{mol C}_2\text{H}_6\text{O}}{46.07 \, \text{g}} \times \frac{2 \, \text{mol CO}_2}{1 \, \text{mol C}_2\text{H}_6\text{O}} = 1.35 \, \text{mol CO}_2 \) 2. **Calculation using O₂:** - Moles of CO₂ formed \( = 31.0 \, \text{g O}_2 \times \frac{1 \, \text{mol O}_2}{32.00 \, \text{g}} \times \frac{2 \, \text{mol CO}_2}{3 \, \text{mol O}_2} = 0.646 \, \text{mol CO}_2 \) **Conclusion:** Since the number of moles produced by O₂ is less, O₂ is the limiting reagent.
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