Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 3 - Chemical Reactions and Reaction Stoichiometry - Exercises - Page 115: 3.36d


3.265x10^22 atoms O

Work Step by Step

First we need to know how many moles of O are present in 1 mole of Al(NO3)3. Looking at the formula we see that there are 9: 3 in each NO3 and there are 3 NO3s so 3*3 = 9. Then, 1 mole of Al(NO3)3 would have 6.022 x 10^23 atoms of Al(NO3)3 so then 6.025x10^-3 mol * (6.022 x 10^23 molecules Al(NO3)3 /1mole)*(9 atoms O/ 1 molecule Al(NO3)3) = 3.265x10^22 atoms O
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