Considering the chemical equation below 4 KO2(s) + 2 H20(1) → 4 KOH(s) + 3 O2(g) When 4.50 g of H20 was.reacted, the percent yield of oxygen was 73.2 %. What was the actual yield of oxygen? a. 2.78 g O b. 5.85 g O c. 3.90 g O d. 8.77 g e. 16.4 g

Considering the chemical equation below 4 KO2(s) + 2 H20(1) → 4 KOH(s) + 3 O2(g) When 4.50 g of H20 was.reacted, the percent yield of oxygen was 73.2 %. What was the actual yield of oxygen? a. 2.78 g O b. 5.85 g O c. 3.90 g O d. 8.77 g e. 16.4 g

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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Concept explainers

Mole Concept

Mole concept is a method to determine the amount of any substance that consists of some elemental particles using a grouping unit called ‘mole’. The mole concept enables one to handle the large amount of small elementary particles. The mole concept can be used in many calculations related to mole calculation, molar amount, mole ratio, and so on, which are significant in chemistry.

Question

Considering the chemical equation below
4 KO2(s) + 2 H2O(1) → 4 KOH(s) + 3 O2(g)
When 4.50 g of H20 was.reacted, the percent yield of oxygen was 73.2 %. What was the actual
yield of oxygen?
O a. 2.78 g
O b. 5.85 g
1ニニ=
ール
c. 3.90 g
O d. 8.77 g
O e. 16.4 g

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