Question

Consider the balanced reactionbelow:2FeBr3 + 3Na2S → Fe2S3 +6NaBrHow many moles of iron(III) sulfide,Fe2S3, would be produced from thecomplete reaction of 449 g iron(III)bromide, FeBr3?

Answer 1

Answers: Moles of Iron (III) Sulphide = 0.76 moles

Explanations :

Given :

• Molar mass of FeBr3 = 295.56 g/mol

,

• Mass of FeBr3 = 449 g

(I) Calculate moles of FeBr3

`\begin{gathered} \text{Moles = }\frac{mass\text{ }}{\text{Molecular mass }} \\ \text{ = }\frac{449\text{ g }}{295.56\text{ g/mol}} \\ \text{ =1.519 } \\ \text{ }\approx1.52molesofFeBr_3 \end{gathered}`

(II) Calculate moles of Fe2S3

The balanced reaction is given as :

`2FeBr_3+3Na_2S\text{ }\Rightarrow Fe_2S_{3\text{ }}+\text{ 6NaBr }`

by stoichiometry , we can see that :

• 2 moles of FeBr3 produces 1 moles of Fe2S3

Therefore ;

• 1.52 mol of febr3 produces X Fe2S3

,

• X moles of FE2S3, = (1.52moles FeBr3 * 1 mol Fe2S3) /2 moles

FeBr3.

=0.755 moles

≈0.76 moles

• This means that 0.76 moles of Fe2S3 would be produced from the complete reaction of 449 g iron(III)bromide,