Solution Found!
Answer: A common laboratory preparation of oxygen gas is
Chapter 3, Problem 78P(choose chapter or problem)
A common laboratory preparation of oxygen gas is the thermal decomposition of potassium chlorate \(\left(\mathrm{KClO}_{3}\right)\). Assuming complete decomposition, calculate the number of grams of \(\mathrm{O}_{2}\) gas that can be obtained from 46.0 g of \(\mathrm{KClO}_{3}\). (The products are KCl and \(\mathrm{O}_{2}\).)
Questions & Answers
(1 Reviews)
QUESTION:
A common laboratory preparation of oxygen gas is the thermal decomposition of potassium chlorate \(\left(\mathrm{KClO}_{3}\right)\). Assuming complete decomposition, calculate the number of grams of \(\mathrm{O}_{2}\) gas that can be obtained from 46.0 g of \(\mathrm{KClO}_{3}\). (The products are KCl and \(\mathrm{O}_{2}\).)
ANSWER:Step 1 of 2
Here we have to calculate the number of grams of \(\mathrm{O}_{2}\) gas that can be obtained from 46.0 g of \(\mathrm{KClO}_{3}\).
The thermal decomposition of potassium chlorate \(\left(\mathrm{KClO}_{3}\right)\) is given below,
\(2\mathrm{KClO}_3(\mathrm{s})\rightarrow2\mathrm{KCl}(\mathrm{s})+3\mathrm{O}_2(\mathrm{g})\)
Given mass of \(\mathrm{KClO}_3=46.0\mathrm{\ g}\)
Molecular mass of \(\mathrm{KClO}_3=122.55\mathrm{\ g}\)
Atomic mass of \(\mathrm{O}_2=32.00\mathrm{\ g}\)
Reviews
Review this written solution for 177619) viewed: 3867 isbn: 9780073402680 | Chemistry - 11 Edition - Chapter 3 - Problem 78p
Thank you for your recent purchase on StudySoup. We invite you to provide a review below, and help us create a better product.
No thanks, I don't want to help other students