22. Be able to solve one-step and multi-step problems involving moles, mass, and a balanced chemical reaction (stoichiometry), such as the problem below, using dimensional analysis. 2 Na + CaBr2 > Ca + 2 NaBr Determine the mass of NaBr that can be formed by the reaction of 1.94 mol CaBr2.
22. Be able to solve one-step and multi-step problems involving moles, mass, and a balanced chemical reaction (stoichiometry), such as the problem below, using dimensional analysis. 2 Na + CaBr2 > Ca + 2 NaBr Determine the mass of NaBr that can be formed by the reaction of 1.94 mol CaBr2.
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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![## Stoichiometry Practice Problem
### Objective:
Be able to solve one-step and multi-step problems involving moles, mass, and a balanced chemical reaction (stoichiometry), using dimensional analysis. An example problem is provided below.
### Example Problem:
Consider the given balanced chemical reaction:
\[ 2 \, \text{Na} + \text{CaBr}_2 \rightarrow \text{Ca} + 2 \, \text{NaBr} \]
Determine the mass of NaBr that can be formed by the reaction of 1.94 mol of CaBr\(_2\).
### Solution:
To solve this problem, follow these steps:
1. **Identify the stoichiometric relationship from the balanced equation.**
For every 1 mole of CaBr\(_2\), 2 moles of NaBr are produced.
2. **Calculate the moles of NaBr produced.**
Given: 1.94 moles of CaBr\(_2\)
Using the stoichiometric ratio:
\[
1.94 \text{ moles of CaBr}_2 \times \frac{2 \text{ moles of NaBr}}{1 \text{ mole of CaBr}_2} = 3.88 \text{ moles of NaBr}
\]
3. **Convert moles of NaBr to mass of NaBr.**
The molar mass of NaBr (sodium bromide):
\[
\text{Molar mass of NaBr} = 22.99 \text{ g/mol (Na)} + 79.90 \text{ g/mol (Br)} = 102.89 \text{ g/mol}
\]
Mass of NaBr:
\[
3.88 \text{ moles of NaBr} \times 102.89 \text{ g/mol} = 399.21 \text{ grams}
\]
### Conclusion:
The mass of NaBr that can be formed from the reaction of 1.94 moles of CaBr\(_2\) is 399.21 grams.
Use this step-by-step guide and similar problems to practice and master stoichiometry calculations.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2b52cd77-b3d2-4b5d-8903-b23362bef087%2F1d819577-1593-4156-9bd9-c1051ea099a5%2Fe3wiyfq_processed.png&w=3840&q=75)
Transcribed Image Text:## Stoichiometry Practice Problem
### Objective:
Be able to solve one-step and multi-step problems involving moles, mass, and a balanced chemical reaction (stoichiometry), using dimensional analysis. An example problem is provided below.
### Example Problem:
Consider the given balanced chemical reaction:
\[ 2 \, \text{Na} + \text{CaBr}_2 \rightarrow \text{Ca} + 2 \, \text{NaBr} \]
Determine the mass of NaBr that can be formed by the reaction of 1.94 mol of CaBr\(_2\).
### Solution:
To solve this problem, follow these steps:
1. **Identify the stoichiometric relationship from the balanced equation.**
For every 1 mole of CaBr\(_2\), 2 moles of NaBr are produced.
2. **Calculate the moles of NaBr produced.**
Given: 1.94 moles of CaBr\(_2\)
Using the stoichiometric ratio:
\[
1.94 \text{ moles of CaBr}_2 \times \frac{2 \text{ moles of NaBr}}{1 \text{ mole of CaBr}_2} = 3.88 \text{ moles of NaBr}
\]
3. **Convert moles of NaBr to mass of NaBr.**
The molar mass of NaBr (sodium bromide):
\[
\text{Molar mass of NaBr} = 22.99 \text{ g/mol (Na)} + 79.90 \text{ g/mol (Br)} = 102.89 \text{ g/mol}
\]
Mass of NaBr:
\[
3.88 \text{ moles of NaBr} \times 102.89 \text{ g/mol} = 399.21 \text{ grams}
\]
### Conclusion:
The mass of NaBr that can be formed from the reaction of 1.94 moles of CaBr\(_2\) is 399.21 grams.
Use this step-by-step guide and similar problems to practice and master stoichiometry calculations.
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