Answer
$50.91\%$
Work Step by Step
Number of moles of NaOH:
$11.1\ mL\times 0.121\ M=1.34\ mmol$, same as excess HCl.
Total number of moles of HCl:
$50.0\ mL\times 0.100\ M=5.00\ mmol$
Number of moles consumed: $5.00-1.34=3.66\ mmol$, same as ammonia.
From stoichiometry, the number of moles of ammonium sulfate: $1.83\ mmol$
Mass: $0.00183\ mol\times 132.14\ g/mol=0.242\ g$
Mass percentage: $0.242/0.475\times100\%=50.91\%$
