Answer
0.244g KCl
Work Step by Step
Moles of AgNO3 in in 15 mL of 0.200 M solution
0.200 m/L * 0.015 L =0.003 mol AgNO3
0.003 mol AgNO3 * (74.55g KCl)/(1 mol) =0.244g KCl
Chemistry: The Central Science (13th Edition)
by Brown, Theodore E.; LeMay, H. Eugene; Bursten, Bruce E.; Murphy, Catherine; Woodward, Patrick; Stoltzfus, Matthew E.
Published by
Prentice Hall
ISBN 10:
0321910419
ISBN 13:
978-0-32191-041-7
Chapter 4 - Reactions in Aqueous Solution - Exercises - Page 160: 4.79b
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