Chemistry 10th Edition

by Whitten, Kenneth W.; Davis, Raymond E.; Peck, Larry; Stanley, George G.

Published by Brooks/Cole Publishing Co.

ISBN 10: 1133610668

ISBN 13: 978-1-13361-066-3

Chapter 3 - Chemical Equations and Reaction of Stoichiometry - Exercises - Mixed Exercises - Page 111: 86

Answer

$m(Br_2)=1.422g$

Work Step by Step

The following reaction takes place: $Cl_2+2KBr\rightarrow Br_2 + 2KCl$ Considering stoichiometric coefficients, we can conclude that $m(Br_2)=m(Cl_2)\times \frac{M(Br_2)}{M(Cl_2)}=0.631g \times \frac{160\frac{g}{mol}}{71\frac{g}{mol}}=1.422g$

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