Answer
0.29 mol KCN
Work Step by Step
$4Au+8KCN+O_{2}+2H_{2}O$ -> $4KAu(CN)_{2}+4KOH$
mol KCN = $\frac{(29.0gAu)\times(1molAu)\times(8molKCN)}{(197gAu)\times(4molAu)}$
$\approx$ 0.2944162437
=0.29 mol KCN
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 110: 3.74
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