**Chemistry Reaction Problem: Calculating Percent Yield****Problem Statement:**You have 2.2 mol Xe and 3.2 mol F₂, but when you carry out the reaction, you end up with only 0.25 mol XeF₄. What is the percent yield of this experiment?**Chemical Reaction:**\[ \text{Xe(g) + 2 F}_2\text{(g) } \rightarrow \text{ XeF}_4\text{(g)} \]**Diagram:**The image appears to have a graphical section with a percentage indicator, but it is not fully visible due to screen damage.**Calculation Steps:**1. **Balanced Reaction Equation:** - Based on the reaction, Xe reacts with F₂ to form XeF₄.2. **Determine the Limiting Reagent:** - Use stoichiometry to find the theoretical yield of XeF₄. - From the reaction, 1 mol Xe reacts with 2 mol F₂. - For 2.2 mol Xe: Needs 4.4 mol F₂ (not enough F₂, so F₂ is the limiting reagent). - For 3.2 mol F₂: Reacts fully with 1.6 mol Xe to produce 1.6 mol XeF₄.3. **Calculate Percent Yield:** - Percent yield = (Actual yield / Theoretical yield) * 100 - Actual yield = 0.25 mol XeF₄ - Theoretical yield = 1.6 mol XeF₄ - Percent yield = (0.25 mol / 1.6 mol) * 100 = 15.63%This problem involves understanding stoichiometry, identifying limiting reagents, and calculating the percent yield based on experimental data.

We have to calculate the percent yield of the given reaction.

Answered: You have 2.2 mol Xe and 3.2 mol F2, but…

You have 2.2 mol Xe and 3.2 mol F2, but when you carry out the reaction you end up with only 0.25 mol XeF4. What is the percent yield of this experiment? Xe(g) + 2 F₂ (g) → XeF. (g)

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

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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Concept explainers

Stoichiometry

Stoichiometry is a chemical division that evaluates the quantitative content of the chemical process with the help of the reacting molecules or products taking part in the reaction. The term “stoicheion” in “Stoichiometry”, is an old Greek term for "element" whereas “metron” means "measure".

Question

**Chemistry Reaction Problem: Calculating Percent Yield**

**Problem Statement:**

You have 2.2 mol Xe and 3.2 mol F₂, but when you carry out the reaction, you end up with only 0.25 mol XeF₄. What is the percent yield of this experiment?

**Chemical Reaction:**

\[ \text{Xe(g) + 2 F}_2\text{(g) } \rightarrow \text{ XeF}_4\text{(g)} \]

**Diagram:**

The image appears to have a graphical section with a percentage indicator, but it is not fully visible due to screen damage.

**Calculation Steps:**

1. **Balanced Reaction Equation:**
- Based on the reaction, Xe reacts with F₂ to form XeF₄.

2. **Determine the Limiting Reagent:**
- Use stoichiometry to find the theoretical yield of XeF₄.
- From the reaction, 1 mol Xe reacts with 2 mol F₂.
- For 2.2 mol Xe: Needs 4.4 mol F₂ (not enough F₂, so F₂ is the limiting reagent).
- For 3.2 mol F₂: Reacts fully with 1.6 mol Xe to produce 1.6 mol XeF₄.

3. **Calculate Percent Yield:**
- Percent yield = (Actual yield / Theoretical yield) * 100
- Actual yield = 0.25 mol XeF₄
- Theoretical yield = 1.6 mol XeF₄
- Percent yield = (0.25 mol / 1.6 mol) * 100 = 15.63%

This problem involves understanding stoichiometry, identifying limiting reagents, and calculating the percent yield based on experimental data.

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