How many grams of the excess reagent remain when 19.65 g of KO2 and 1.95 g of H20react according to the balanced chemical equation below? You may assume 100% yield forthe reaction.Molar masses (in g/mol):КО2 3D 71.1H20 = 18.02КОН56.11О2 %3D 32.004КО2 + 2H20} 302 + 4KОНO 7.69 g of excess KO24.26 g of excess KO28.84 g of excess O2O 15.5 g of excess KOH1.08 g of excess H204.98 gof excess H2O

Here we are calculating excess reagent

How many grams of the excess reagent remain when 19.65 g of KO2 and 1.95 g of H20 react according to the balanced chemical equation below? You may assume 100% yield for the reaction. Molar masses (in g/mol): КО, 3 71.1 H20 = 18.02 KOH = 56.11 O2 = 32.00 4КО2 + 2H20 > 302 + 4КОН 7.69 g of excess KO2 4.26 g of excess KO2 8.84 g of excess O2 15.5 g of excess KOH 1.08 g of excess H20 4.98 g of excess H20

How many grams of the excess reagent remain when 19.65 g of KO2 and 1.95 g of H20 react according to the balanced chemical equation below? You may assume 100% yield for the reaction. Molar masses (in g/mol): КО, 3 71.1 H20 = 18.02 KOH = 56.11 O2 = 32.00 4КО2 + 2H20 > 302 + 4КОН 7.69 g of excess KO2 4.26 g of excess KO2 8.84 g of excess O2 15.5 g of excess KOH 1.08 g of excess H20 4.98 g of excess H20

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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