# How many moles of iron (III) sulfide are produced from the reaction of 449 grams of iron (III) bromide?

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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

Formulae;

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

Math:

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.

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Not sure what happened to my math, but 449/295.645 is 1.5187, not 1.8278, so the final answer is 1.5187/2 = .759 mol.

This is a Stoichiometry problem since it involves calculations in chemical changes. The first step of all stoichiometry problems is to write a balanced equation and define the given and identify what you are seeking. The problem identifies iron(III) bromide as a reactant and iron(III) sulfide as a product. However it doesn't identify the other reactant. The identity of the other reactant does not matter, as long as it has the sulfide ion and is in excess. (This means that the iron(III) bromide runs out and stops the production of iron(III) sulfide.)

So you may choose any compound that contains sulfide, for example, sodium sulfide, and write the balanced equation:

2 FeBr3 + 3 Na2S -----> 6 NaBr + Fe2S3

449 g. ? Mol

The coefficients in the balanced equation represent moles so the equation indicates that 1 mole of Fe2S3 is produced for every 2 moles of FeBr3. So if you can convert the 449 g of FeBr3 to moles, you can then use the mole ratio stated above. To convert mass to moles, use the molar mass which is 295.55 g/mol.

449 g FeBr3 x 1 mole FeBr3/295.55 g FeBr3 = 1.52 mole FeBr3

Next use the mole ratio from the balanced equation:

1.52 mole FeBr3 x 1 mole Fe2S3/2 mole FeBr3 **= 0.760 moles Fe2S3**

This answer uses 3 significant figures which are rounded at the end of the calculation.

The balanced reaction is

2FeBr₃ + 3H₂S `->` Fe₂S₃ + 6HBr

Molar mass of Iron(III) Sulfide is 207.9 g/mol

Molar mass of Iron(III) Bromide is 295.56 g/mol

Therefore 2*295.56 g of Iron(III) Bromide produces 207.9 g of Iron(III) Sulfide

`:.` 449 grams of Iron(III) Bromide produces (207.9*449)/(2*295.56) grams of Iron(III) Sulfide .` `

Since Molar mass of Iron(III) Sulfide is 207.9 ,

Number of moles of Iron(III) Sulfide produced = 449/(2*295.56)

= 449/591.12 = **0.76**