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

(II) Calculate moles of Fe2S3
The balanced reaction is given as :

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,