Hydrogen peroxide can be prepared by the reaction of barium peroxide with sulfuric acid according to the reaction:\[ \text{BaO}_2(s) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{BaSO}_4(s) + \text{H}_2\text{O}_2(aq) \]How many milliliters of 3.75 M \(\text{H}_2\text{SO}_4(aq)\) are needed to react completely with 62.3 g \(\text{BaO}_2(s)\)?\[ V = \, \_\_\_\_\_\_\_\_\_\_\, \text{mL} \]There are no graphs or diagrams provided within this content.

Given, Mass of BaO2 is = 62.3 g Molar mass of BaO2 is = 169.33 g/mol Moles of BaO2 = mass/ molar…

Hydrogen peroxide can be prepared by the reaction of barium peroxide with sulfuric acid according to the reaction BaO₂ (s) + H₂SO4 (aq) → -> BaSO4(s) + H₂O₂(aq) How many milliliters of 3.75 M H₂SO4 (aq) are needed to react completely with 62.3 g BaO₂ (s)? V = mL

Hydrogen peroxide can be prepared by the reaction of barium peroxide with sulfuric acid according to the reaction BaO₂ (s) + H₂SO4 (aq) → -> BaSO4(s) + H₂O₂(aq) How many milliliters of 3.75 M H₂SO4 (aq) are needed to react completely with 62.3 g BaO₂ (s)? V = mL

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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Question

Hydrogen peroxide can be prepared by the reaction of barium peroxide with sulfuric acid according to the reaction:

\[ \text{BaO}_2(s) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{BaSO}_4(s) + \text{H}_2\text{O}_2(aq) \]

How many milliliters of 3.75 M \(\text{H}_2\text{SO}_4(aq)\) are needed to react completely with 62.3 g \(\text{BaO}_2(s)\)?

\[ V = \, \_\_\_\_\_\_\_\_\_\_\, \text{mL} \]

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