Determine the limiting reactant, given 31.0 g of each reactant, in the following: C,H6O(1) + 302 (g) 2CO2 (g) + 3H2O(g)

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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**Part H: Calculating the Mass of H₂O Produced**

You are given the task to calculate the grams of H₂O (water) that would be produced in a chemical reaction. Ensure that your answer is expressed with the appropriate units.

**Input Section:**

- You will find a text box labeled "m(H₂O) =" where you need to enter:
  - **Value:** The numerical result of your calculation.
  - **Units:** The units of measurement for your answer (e.g., grams).

**Interactive Tools:**

- The interface provides tools for entering mathematical symbols and units. These include options for scientific notation, Greek letters, and unit symbols.

**Options:**

- **Submit:** Click here when ready to submit your answer.
- **Request Answer:** Use this if you need guidance or assistance.

**Feedback and Navigation:**

- **Provide Feedback:** Option available for submitting feedback on the problem.
- **Next:** Button to navigate to the subsequent part of the activity.

**Branding:**

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This exercise helps reinforce skills in stoichiometry and unit conversion, which are crucial in chemistry.
Transcribed Image Text:**Part H: Calculating the Mass of H₂O Produced** You are given the task to calculate the grams of H₂O (water) that would be produced in a chemical reaction. Ensure that your answer is expressed with the appropriate units. **Input Section:** - You will find a text box labeled "m(H₂O) =" where you need to enter: - **Value:** The numerical result of your calculation. - **Units:** The units of measurement for your answer (e.g., grams). **Interactive Tools:** - The interface provides tools for entering mathematical symbols and units. These include options for scientific notation, Greek letters, and unit symbols. **Options:** - **Submit:** Click here when ready to submit your answer. - **Request Answer:** Use this if you need guidance or assistance. **Feedback and Navigation:** - **Provide Feedback:** Option available for submitting feedback on the problem. - **Next:** Button to navigate to the subsequent part of the activity. **Branding:** - The bottom of the page is branded with "Pearson," indicating the educational provider of this content. This exercise helps reinforce skills in stoichiometry and unit conversion, which are crucial in chemistry.
**Limiting Reactant Determination**

To determine the limiting reactant, given 31.0 g of each reactant in the following chemical reaction:

\[ \text{C}_2\text{H}_6\text{O(l)} + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O(g)} \]

### Possible Reactants:
- \(\text{H}_2\text{O}\)
- \(\text{C}_2\text{H}_6\text{O}\)
- \(\text{O}_2\)
- \(\text{CO}_2\)

### Solution:

**Step-by-Step Calculation:**

1. **Moles of CO\(_2\) from C\(_2\)H\(_6\)O:**

   \[
   \text{moles of CO}_2 = 31.0 \, \text{g} \, \text{C}_2\text{H}_6\text{O} \times \frac{1 \, \text{mol} \, \text{C}_2\text{H}_6\text{O}}{46.07 \, \text{g}} \times \frac{2 \, \text{mol} \, \text{CO}_2}{1 \, \text{mol} \, \text{C}_2\text{H}_6\text{O}} = 1.35 \, \text{mol CO}_2
   \]

2. **Moles of CO\(_2\) from O\(_2\):**

   \[
   \text{moles of CO}_2 = 31.0 \, \text{g} \, \text{O}_2 \times \frac{1 \, \text{mol} \, \text{O}_2}{32.00 \, \text{g}} \times \frac{2 \, \text{mol} \, \text{CO}_2}{3 \, \text{mol} \, \text{O}_2} = 0.646 \, \text{mol CO}_2
   \]

### Conclusion:
Since the number of moles produced by O\(_2\)
Transcribed Image Text:**Limiting Reactant Determination** To determine the limiting reactant, given 31.0 g of each reactant in the following chemical reaction: \[ \text{C}_2\text{H}_6\text{O(l)} + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O(g)} \] ### Possible Reactants: - \(\text{H}_2\text{O}\) - \(\text{C}_2\text{H}_6\text{O}\) - \(\text{O}_2\) - \(\text{CO}_2\) ### Solution: **Step-by-Step Calculation:** 1. **Moles of CO\(_2\) from C\(_2\)H\(_6\)O:** \[ \text{moles of CO}_2 = 31.0 \, \text{g} \, \text{C}_2\text{H}_6\text{O} \times \frac{1 \, \text{mol} \, \text{C}_2\text{H}_6\text{O}}{46.07 \, \text{g}} \times \frac{2 \, \text{mol} \, \text{CO}_2}{1 \, \text{mol} \, \text{C}_2\text{H}_6\text{O}} = 1.35 \, \text{mol CO}_2 \] 2. **Moles of CO\(_2\) from O\(_2\):** \[ \text{moles of CO}_2 = 31.0 \, \text{g} \, \text{O}_2 \times \frac{1 \, \text{mol} \, \text{O}_2}{32.00 \, \text{g}} \times \frac{2 \, \text{mol} \, \text{CO}_2}{3 \, \text{mol} \, \text{O}_2} = 0.646 \, \text{mol CO}_2 \] ### Conclusion: Since the number of moles produced by O\(_2\)
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