Answer:
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.
