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.

the answer

explanation:

the capillary tube was too close to the bottom of the beaker.

the ruler may have moved.

water got into the capillary tube.

the temperature was not allowed to equilibrate in the 2-4 minutes.