Chemistry 12th Edition

Published by McGraw-Hill Education
ISBN 10: 0078021510
ISBN 13: 978-0-07802-151-0

Chapter 3 - Mass Relationships in Chemical Reactions - Questions & Problems: 3.76

Answer

a. $NH_{4}NO_{3}$ -> $N_{2}O+2H_{2}O$ b. $20g$ $N_{2}O$

Work Step by Step

a. $NH_{4}NO_{3}$ -> $N_{2}O+2H_{2}O$ b. mass of $N_{2}O$ = $\frac{(0.46molNH_{4}NO_{3})\times(1molN_{2}O)\times(44.02gN_{2}O)}{(1molNH_{4}NO_{3})\times(1molN_{2}O)}$ $=20g$ $N_{2}O$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.