Answer
$171.6\ L$
Work Step by Step
The number of moles of gold: $1000\ kg\times 1000\ g/kg\times 0.012/100\div 196.97\ g/mol=0.61\ mol$
Number of moles of NaCN: $0.61\ mol\times 8/4=1.22\ mol$
Volume: $1.22\ mol\div 0.0071\ M=171.6\ L$
Chemistry and Chemical Reactivity (9th Edition)
by Kotz, John C.; Treichel, Paul M.; Townsend, John R.; Treichel, David A.
Published by
Cengage Learning
ISBN 10:
1133949649
ISBN 13:
978-1-13394-964-0
Chapter 17 Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria - 17-7 Solubility and Complex Ions - Applying Chemical Principles - Questions - Page 675: 2
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