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Solved: An emergency breathing apparatus placed in mines or caves works via the chemical

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro ISBN: 9780321910295 34

Solution for problem 96P Chapter 8

Introductory Chemistry | 5th Edition

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Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro

Introductory Chemistry | 5th Edition

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1
Problem 96P

An emergency breathing apparatus placed in mines or caves works via the chemical reaction:

\(4 \mathrm{KO}_{2}(s)+2 \mathrm{CO}_{2}(g) \rightarrow 2 \mathrm{~K}_{2} \mathrm{CO}_{3}(s)+3 \mathrm{O}_{2}(g)\)

If the oxygen supply becomes limited or if the air becomes poisoned, a worker can use the apparatus to breathe while exiting the mine. Notice that the reaction produces \(\mathrm{O}_{2}\), which can be breathed, and absorbs \(\mathrm{CO}_{2}\), a product of respiration. What minimum amount of \(\mathrm{KO}_{2}\) is required for the apparatus to produce enough oxygen to allow the user 15 minutes to exit the mine in an emergency? Assume that an adult consumes approximately 4.4 g of oxygen in 15 minutes of normal breathing.

Equation Transcription:

Text Transcription:

4 KO_2 (s) + 2 CO_2 (g) right arrow 2 K_2 CO_3 (s) + 3 O_2 (g)

O_2

CO_2

KO_2

Step-by-Step Solution:
Step 1 of 3

Solution 96P

4 KO2(s) + 2 CO2(g)→2 K2CO3(s) + 3 O2(g)

4 moles K2O2 react to produce 3 moles of O2

moles of O2: 5.6 g/32 g/mole O2=0.175

0.175 moles O2 x 4/3 x 71.1 g/mole K2O=16.6 g K2O required.

Step 2 of 3

Chapter 8, Problem 96P is Solved
Step 3 of 3

Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

The full step-by-step solution to problem: 96P from chapter: 8 was answered by , our top Chemistry solution expert on 05/06/17, 06:45PM. Introductory Chemistry was written by and is associated to the ISBN: 9780321910295. This full solution covers the following key subjects: apparatus, oxygen, mine, minutes, reaction. This expansive textbook survival guide covers 19 chapters, and 2046 solutions. The answer to “?An emergency breathing apparatus placed in mines or caves works via the chemical reaction:\(4 \mathrm{KO}_{2}(s)+2 \mathrm{CO}_{2}(g) \rightarrow 2 \mathrm{~K}_{2} \mathrm{CO}_{3}(s)+3 \mathrm{O}_{2}(g)\)If the oxygen supply becomes limited or if the air becomes poisoned, a worker can use the apparatus to breathe while exiting the mine. Notice that the reaction produces \(\mathrm{O}_{2}\), which can be breathed, and absorbs \(\mathrm{CO}_{2}\), a product of respiration. What minimum amount of \(\mathrm{KO}_{2}\) is required for the apparatus to produce enough oxygen to allow the user 15 minutes to exit the mine in an emergency? Assume that an adult consumes approximately 4.4 g of oxygen in 15 minutes of normal breathing.Equation Transcription:Text Transcription:4 KO_2 (s) + 2 CO_2 (g) right arrow 2 K_2 CO_3 (s) + 3 O_2 (g)O_2CO_2KO_2” is broken down into a number of easy to follow steps, and 122 words. Since the solution to 96P from 8 chapter was answered, more than 671 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Introductory Chemistry, edition: 5.

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Solved: An emergency breathing apparatus placed in mines or caves works via the chemical