**Problem: Percent Yield Calculation in Chromium (III) Oxide Reaction**59.6 g of chromium (III) oxide were formed from the reaction of 25.0 g of O₂ and excess chromium metal. What was the percent yield of this reaction? The formula mass of Cr₂O₃ is 156.0 g/mol.**Chemical Equation:**\[ 4 \text{Cr}(s) + 3 \text{O}_2(g) \rightarrow 2 \text{Cr}_2\text{O}_3(s) \]**Instructions:**Enter a numerical value for your answer, omitting the units of %. Your answer should have three significant figures.Do NOT enter the decimal form of the percentage.

Answered: Image /qna-images/answer/e160a38d-94ca-42fb-af5e-489e15878043.jpg

59.6 g of chromium (III) oxide were formed from reaction of 25.0 g of O₂ and excess chromium metal. What was the percent yield of this reaction. The formula mass of Cr₂O3 is 156.0 g/mol. 4 Cr(s) + 3 O₂(g) → 2 Cr₂O3(s) Enter a numerical value for your answer; omit the units of %. Your answer should have three significant figures. Do NOT enter the decimal form of the percentage.

59.6 g of chromium (III) oxide were formed from reaction of 25.0 g of O₂ and excess chromium metal. What was the percent yield of this reaction. The formula mass of Cr₂O3 is 156.0 g/mol. 4 Cr(s) + 3 O₂(g) → 2 Cr₂O3(s) Enter a numerical value for your answer; omit the units of %. Your answer should have three significant figures. Do NOT enter the decimal form of the percentage.

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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