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 - Page 107: 3.22


Please see the work below.

Work Step by Step

First of all, we find the number of moles for 2 atoms of Pb and then convert it into mass $2\space atoms \space Pb\times\frac{1mol}{6.022\times 10^{23}atoms}=3.321\times 10^{-24}mol\space Pb$ $3.321\times 10^{-24} mol \space Pb\times \frac{207.2g\space Pb}{1mol\space Pb}=6.881\times 10^{-22}g\space Pb$ Now we convert $5.1\times 10^{-23}$ mol of He to mass $5.1\times 10^{-23}mol\space He \times \frac{4.003g\space He}{1mol\space He}=2.0\times 10^{-22}g\space He$ Thus 2 atoms of lead has greater mass than $5.1\times 10^{-23} mol$ of He.
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