How many moles of iron(III) sulfide, Fe2S3, would be produced from the

complete reaction of 449 g iron(III)

bromide, FeBr3?

0.76 mole of Fe2S3.

Explanation:

Step 1:

Determination of the number of mole in 449g iron(III)bromide, FeBr3. This is illustrated below:

Mass of FeBr3 = 449g

Molar mass of FeBr3 = 56 + (80x3) = 296g/mol

Mole of FeBr3 =..?

Mole = Mass /Molar Mass

Mole of FeBr3 = 449/296

Mole of FeBr3 = 1.52 moles

Step 2:

The balanced equation for the reaction. This is given below:

2FeBr3 + 3Na2S —> 6NaBr + Fe2S3

Step 3:

Determination of the number of mole of Fe2S3 produced from the reaction of 449g ( i.e 1.52 moles) of FeBr3. This is illustrated below:

From the balanced equation above,

2 moles of FeBr3 reacted to produce 1 mole of Fe2S3.

Therefore, 1.52 moles of FeBr3 will react to produce = (1.52 x 1)/2 = 0.76 mole of Fe2S3.

Therefore, 0.76 mole of Fe2S3 is produced from the reaction.