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.
$0.035\ mol\times 439.93\ g/mol=15.40\ g$