Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 4 Stoichiometry: Quantitative Information about Chemical Reactions - Study Questions - Page 179h: 100

Answer

$15.40\ g$

Work Step by Step

Number of moles of Ag: $8.63\ g\div 107.87\ g/mol=0.080\ mol$ $0.040\ mol\ Ag_2MoS_4$ Number of moles of Mo: $3.36\ g\div 95.95\ g/mol=0.035\ mol$ $0.035\ mol\ Ag_2MoS_4$ Number of moles of S: $4.81\ g\div 32.06\ g/mol=0.150\ mol$ $0.038\ mol\ Ag_2MoS_4$ Molybdenum is the limiting reactant. $0.035\ mol\times 439.93\ g/mol=15.40\ g$
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