Bombardier beetles inject boiling hydrogen peroxide, or H,O,, into their victims. It then decomposes into hydrogen and oxygen gas. How many molecules of oxygen can be produced if 4.10 moles of hydrogen peroxide are decomposed?
Bombardier beetles inject boiling hydrogen peroxide, or H,O,, into their victims. It then decomposes into hydrogen and oxygen gas. How many molecules of oxygen can be produced if 4.10 moles of hydrogen peroxide are decomposed?
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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![**Educational Content: Chemistry**
**Topic: Chemical Reactions and Stoichiometry**
**Example Problem: Decomposition of Hydrogen Peroxide**
*Problem Statement:*
Bombardier beetles inject boiling hydrogen peroxide, or \( H_2O_2 \), into their victims. It then decomposes into hydrogen and oxygen gas. How many molecules of oxygen can be produced if 4.10 moles of hydrogen peroxide are decomposed?
*Solution:*
First, we need to understand the decomposition reaction of hydrogen peroxide. The balanced chemical equation for the decomposition of hydrogen peroxide is:
\[ 2H_2O_2 \rightarrow 2H_2O + O_2 \]
This equation tells us that 2 moles of \( H_2O_2 \) decomposes to produce 1 mole of \( O_2 \).
1. Find the number of moles of \( O_2 \) produced from 4.10 moles of \( H_2O_2 \):
\[
\frac{1 \text{ mole } O_2}{2 \text{ moles } H_2O_2} \times 4.10 \text{ moles } H_2O_2 = \frac{4.10}{2} \text{ moles } O_2 = 2.05 \text{ moles } O_2
\]
2. Convert moles of \( O_2 \) to molecules of \( O_2 \) using Avogadro's number (\( 6.022 \times 10^{23} \text{ molecules/mole} \)):
\[
2.05 \text{ moles } O_2 \times 6.022 \times 10^{23} \text{ molecules/mole} = 1.23451 \times 10^{24} \text{ molecules}
\]
Therefore, 4.10 moles of hydrogen peroxide can produce approximately \( 1.23 \times 10^{24} \) molecules of oxygen.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd909c482-4f4e-4400-a93e-251a85b7679b%2Fa07a5262-d243-4a05-a14c-2a3474140e6f%2Fn2u497c_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Educational Content: Chemistry**
**Topic: Chemical Reactions and Stoichiometry**
**Example Problem: Decomposition of Hydrogen Peroxide**
*Problem Statement:*
Bombardier beetles inject boiling hydrogen peroxide, or \( H_2O_2 \), into their victims. It then decomposes into hydrogen and oxygen gas. How many molecules of oxygen can be produced if 4.10 moles of hydrogen peroxide are decomposed?
*Solution:*
First, we need to understand the decomposition reaction of hydrogen peroxide. The balanced chemical equation for the decomposition of hydrogen peroxide is:
\[ 2H_2O_2 \rightarrow 2H_2O + O_2 \]
This equation tells us that 2 moles of \( H_2O_2 \) decomposes to produce 1 mole of \( O_2 \).
1. Find the number of moles of \( O_2 \) produced from 4.10 moles of \( H_2O_2 \):
\[
\frac{1 \text{ mole } O_2}{2 \text{ moles } H_2O_2} \times 4.10 \text{ moles } H_2O_2 = \frac{4.10}{2} \text{ moles } O_2 = 2.05 \text{ moles } O_2
\]
2. Convert moles of \( O_2 \) to molecules of \( O_2 \) using Avogadro's number (\( 6.022 \times 10^{23} \text{ molecules/mole} \)):
\[
2.05 \text{ moles } O_2 \times 6.022 \times 10^{23} \text{ molecules/mole} = 1.23451 \times 10^{24} \text{ molecules}
\]
Therefore, 4.10 moles of hydrogen peroxide can produce approximately \( 1.23 \times 10^{24} \) molecules of oxygen.
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