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: 3.66

Answer

1.01 mol $Cl_{2}$

Work Step by Step

$Si(s)+2Cl_{2}(g)$ -> $SiCl_{4}(l)$ mol of $Cl_{@}$= $\frac{(0.507molSiCl_{4})\times(2molCl_{2})}{1molSiCl_{4}}$ = 1.01 mol $Cl_{2}$
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