In problems of stoichiometry I find that it helps to define the terms of the question.
We have a given mass (449g) composed entirely of the same type of molecule, iron(III) bromide. We could split these molecules apart, isolate the iron(III), and recombine it with sulfur, creating iron(III) sulfide and ignoring the leftover bromine. This question is basically asking is how many molecules of this iron(III) sulfide product we could make with the available ingredients.
It also helps to remember that "moles" are like mathematical slang, similar to the number pi. It's just an easy way for us to work with inconvenient numbers. There's nothing intrinsically special about the number 6.022x10e23.
There are several relationships we need to establish in order to solve this problem;
-Molecular formula for both compounds
-mass to moles for iron(III)bromide
-moles to moles for iron(III)bromide and iron(III) sulfide
Iron(III) tells us that this is the iron ion with a +3 charge. Bromine, a halogen, will normally have a -1 charge. Sulfur will be -2. Molecular formulae are FeBr(3) and Fe(2)S(3)
We can already tell that we will need to split two molecules of iron bromide to make one molecule of iron sulfide. This means the molar ratio is 1:2
Molar mass of iron bromide (using a periodic table): 55.933 + (3)79.904 = 295.645
449g iron bromide / 295.645g/mol = 1.8278 mol iron bromide
1.8278 mol iron bromide x (1 mol iron sulfide / 2 mol iron bromide) = .9139 mol iron sulfide
So, about 0.9 moles of iron sulfide can be produced.