Chemistry: The Molecular Nature of Matter and Change 7th Edition

Published by McGraw-Hill Education
ISBN 10: 007351117X
ISBN 13: 978-0-07351-117-7

Chapter 5 - Problems: 5.54

Answer

1.5925 g

Work Step by Step

1 752 torr = 0.989 atm 638 ml = 0.638 liter 128°C = 401°K Convert the given information 2 nO2 = PV/RT = (0.989*0.638)/((22,4/273)*401) = 0.019 mol Rearrange PV=nRT and plug in 3 (0.019*2)/3 = 0.013 mol the ratio between KClO3 and O2 is 2 : 3 4 0.013*122.5 1 mol KClO3 = 122.55 g
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