**Problem Statement:**Given that 10.0 moles of O₂ are reacted with excess NO in the reaction below, and only 6.2 moles of NO₂ were collected, calculate the percent yield for the reaction:\[ 2 \, \text{NO (g)} + \text{O}_2 \, \text{(g)} \rightarrow 2 \, \text{NO}_2 \, \text{(g)} \] **Explanation:**1. **Balanced Chemical Equation:** - 2 moles of NO react with 1 mole of O₂ to produce 2 moles of NO₂.2. **Calculation of Theoretical Yield:** - Since 10.0 moles of O₂ were used, and using the stoichiometry from the balanced equation, 10.0 moles of O₂ would theoretically produce 20.0 moles of NO₂ (as 2 NO₂ are produced per 1 O₂).3. **Actual Yield:** - The actual yield of NO₂ collected is 6.2 moles.4. **Percent Yield Calculation:** - Percent yield is calculated using the formula: \[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% \] - Substitute in the actual yield (6.2 moles) and the theoretical yield (20.0 moles) to get: \[ \text{Percent Yield} = \left( \frac{6.2}{20.0} \right) \times 100\% = 31\% \]Thus, the percent yield of the reaction is 31%. You have 2.2 mol Xe and 2.5 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?\[ \text{Xe(g)} + 2 \text{F}_{2} \text{(g)} \rightarrow \text{XeF}_{4} \text{(g)} \]

The given problem is based on the stoichiometric calculations of the chemical reactions. The given balanced chemical equations are 2 NO (g) + O2 (g) → 2NO2 (g) 2.0 mol 1.0 mol 2.0 mol Given that 10.0 moles of O2 reacted with excess of NO in the given reaction. From the balanced equation, 1.0 mole of O2 gas produces 2.0 moles of NO2 gas Therefore,10.0 mole of O2 gas produces 20.0 moles of NO2 gas The number of moles of NO2 obtained in the reaction = 6.2 moles The percentage yield of the reaction = experimental number of moles of NO2theoretical number of moles of NO2×100 = 6.220.0×100 = 31.0 % Ans: The percentage yield of the reaction = 31.0 % The given balanced chemical equations are Xe (g) + 2F2 (g) → 2XeF4 (g) 1.0 mol 2.0 mol 2.0 mol Given that the number of moles of F2 = 2.5 moles The number of moles of Xe = 2.2 moles From the balanced equation, 1.0 mole of Xe gas reacts with 2.0 moles of F2 gas Therefore, 2.2 moles of Xe gas react with 4.4 moles of F2 gas F2 gas is the limiting reagent and the amount of XeF4 (g) formed depends on the limiting reagent only. From the balanced equation, 2.0 mole of F2 gas produces 2.0 moles of XeF4 gas Therefore, 2.5 mole of F2 gas produces 2.5 moles of XeF4 gas The number of moles of XeF4 gas obtained in the reaction = 0.25 moles The percentage yield of the reaction = experimental number of moles of XeF4theoretical number of moles of XeF4×100 = 0.252.5×100 = 10.0 % Ans: The percentage yield of the reaction = 10.0 %

Answered: If 10.0 moles of O₂ are reacted with…

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Chemistry

If 10.0 moles of O₂ are reacted with excess NO in the reaction below, and only 6.2 mol of NO were collected, then what is the percent yield for the reaction? 2 2 NO (g) + O₂(g) → 2 NO₂ (g) 2

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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