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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![**Chemical Reaction and Percent Yield Calculation**
**Problem Statement:**
A reaction of 49.3 g of Na (sodium) and 41.1 g of Br₂ (bromine) yields 43.0 g of NaBr (sodium bromide). What is the percent yield?
**Chemical Equation:**
\[ 2 \text{Na (s)} + \text{Br}_2 \text{(g)} \rightarrow 2 \text{NaBr (s)} \]
**Calculation Required:**
Determine the percent yield for the reaction.
**Percent Yield Formula:**
\[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% \]
**Input Box:**
There is an input box provided where you can enter the calculated percent yield value.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4aaa6e31-74a0-490c-8490-469c7e8cf5d8%2Fafd590b3-2744-4073-b13c-bf06bdd0b722%2Fd1th9bm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Chemical Reaction and Percent Yield Calculation**
**Problem Statement:**
A reaction of 49.3 g of Na (sodium) and 41.1 g of Br₂ (bromine) yields 43.0 g of NaBr (sodium bromide). What is the percent yield?
**Chemical Equation:**
\[ 2 \text{Na (s)} + \text{Br}_2 \text{(g)} \rightarrow 2 \text{NaBr (s)} \]
**Calculation Required:**
Determine the percent yield for the reaction.
**Percent Yield Formula:**
\[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% \]
**Input Box:**
There is an input box provided where you can enter the calculated percent yield value.
Expert Solution
Step 1
The reaction is as follows:
2Na + Br2 --------->2NaBr.
The mass of Na used in the reaction = 49.3 g
The mass of Br2 used in the reaction = 41.1 g.
The actual yield of NaBr = 43.0 g.
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