Question

How many moles of iron(lll) sulfide, Fe2S3, would be produced from the complete reaction of 449 g iron(lll) bromide, FeBr3?

Answer 1

Answer:

158.004 Grams of Fe2S3

Explanation:

iron(lll) sulfide = Fe2S3 = Molar mass: 207.9 g/mol

iron(lll) bromide = FeBr3 = Molar mass: 295.56 g/mol

Since we don't have the chemical equation we will have to make one I guess.

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

449 g iron(lll) bromide = FeBr3

So we have 449/295.56 = moles of FeBr3 = 1.52 Moles

So ratio is 2:1 FeBr3 to Fe2S3

So we need 0.76 moles of Fe2S3

0.76 x 207.9 = 158.004Grams