Introductory Chemistry (5th Edition)

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

Chapter 15 - Chemical Equilibrium - Exercises - Problems - Page 567: 57

Answer

7.0 M

Work Step by Step

$K_{eq}$ = $\frac{[SbCl_{3}][Cl_{2}]}{[SbCl_{5}]}$ From the above equation, we get $[SbCl_{5}] = \frac{[SbCl_{3}][Cl_{2}]}{K_{eq}}$ = $\frac{0.0255\times0.135}{4.9\times10^{-4}}$ = 7.0 M ( at $248^{\circ}$ C)
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