Introductory Chemistry (5th Edition)

Published by Pearson
ISBN 10: 032191029X
ISBN 13: 978-0-32191-029-5

Chapter 8 - Quantities in Chemical Reactions - Exercises - Problems - Page 274: 23

Answer

(a) 2.4 mol $PbO$, 2.4 mol $SO_{2}$ (b) 1.6 mol $PbO$, 1.6 mol $SO_{2}$ (c) 5.3 mol $PbO$, 5.3 mol $SO_{2}$ (d) 3.5 mol $PbO$, 3.5 mol $SO_{2}$

Work Step by Step

(a) $2.4\,mol\,PbS\times\frac{2\,mol\,PbO}{2\,mol\,PbS}=2.4\,mol\,PbO$ $2.4\,mol\,PbS\times\frac{2\,mol\,SO_{2}}{2\,mol\,PbS}=2.4\,mol\,SO_{2}$ (b) $2.4\,mol\,O_{2}\times\frac{2\,mol\,PbO}{3\,mol\,O_{2}}=1.6\,mol\,PbO$ $2.4\,mol\,O_{2}\times\frac{2\,mol\,SO_{2}}{3\,mol\,O_{2}}=1.6\,mol\,SO_{2}$ (c) $5.3\,mol\,PbS\times\frac{2\,mol\,PbO}{2\,mol\,PbS}=5.3\,mol\,PbO$ $5.3\,mol\,PbS\times\frac{2\,mol\,SO_{2}}{2\,mol\,PbS}=5.3\,mol\,SO_{2}$ (d) $5.3\,mol\,O_{2}\times\frac{2\,mol\,PbO}{3\,mol\,O_{2}}=3.5\,mol\,PbO$ $5.3\,mol\,O_{2}\times\frac{2\,mol\,SO_{2}}{3\,mol\,O_{2}}=3.5\,mol\,SO_{2}$
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