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 $Cl_{2}$
Chemistry 12th Edition
by Chang, Raymond; Goldsby, Kenneth
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 109: 3.66
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