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The reaction between potassium superoxide (KO2, 71.1 g/mol) and CO2 is used as a source of oxygen in self-contained breathing equipment used by rescue workers: 4 KO2 (s) + 2 CO2 (g) -> 2 K2 CO3 (s) + 3 O2 (g) If the consumption of 7.50 g of KO2 in the presence of excess CO 2 results in the production of 0.065 moles of oxygen gas, what is the percent yield of this reaction?

The reaction between potassium superoxide (KO2, 71.1 g/mol) and CO2 is used as a source of oxygen in self-contained breathing equipment used by rescue workers: 4 KO2 (s) + 2 CO2 (g) -> 2 K2 CO3 (s) + 3 O2 (g) If the consumption of 7.50 g of KO2 in the presence of excess CO 2 results in the production of 0.065 moles of oxygen gas, what is the percent yield of this reaction?

Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter4: Reactions In Aqueous Solution
Section: Chapter Questions
Problem 57QAP: A sample of limestone weighing 1.005 g is dissolved in 75.00 mL of 0.2500 M hydrochloric acid. The...
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The reaction between potassium superoxide (KO2, 71.1 g/mol) and CO2 is used as a source of oxygen in self- contained breathing equipment used by rescue workers: 4 KO2 (s) + 2 CO2 (g) -> 2 K2 CO3 (s) + 3 02 (g) If the consumption of 7.50 g of KO2 in the presence of excess CO 2 results in the production of 0.065 moles of oxygen gas, what is the percent yield of this reaction?
Transcribed Image Text:The reaction between potassium superoxide (KO2, 71.1 g/mol) and CO2 is used as a source of oxygen in self- contained breathing equipment used by rescue workers: 4 KO2 (s) + 2 CO2 (g) -> 2 K2 CO3 (s) + 3 02 (g) If the consumption of 7.50 g of KO2 in the presence of excess CO 2 results in the production of 0.065 moles of oxygen gas, what is the percent yield of this reaction?
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