## Chemistry 12th Edition

Published by McGraw-Hill Education

# Chapter 3 - Mass Relationships in Chemical Reactions - Questions & Problems: 3.74

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

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