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Predict the mass of iron (III) sulfide produced when 3.0 g of iron filings react completely with 2.5 g of yellow sulfur solid, S8(s).
A) 1.5g Fe2S3
B) 2.5g Fe2S3
C) 3.5g Fe2S3
D) 5.5g Fe2S3

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superman1987

Answer:

D) 5.5 g Fe₂S₃.

Explanation:

  • The balanced chemical reaction is: 16Fe(s) + 3S₈(s) → 8Fe₂S₃(s)
  • It is clear that 16.0 mole of Fe reacts completely with 3.0 moles of S₈ to produce 8.0 moles of Fe₂S₃.
  • We need to calculate the no. of moles of the reacted Fe filings (3.0 g) and yellow sulfur solid (2.5 g).

No. of moles of Fe filings (3.0 g) = mass /atomic mass = (3.0 g) / (55.845 g/mol) = 0.0537 mol.

No. of moles of yellow sulfur solid (2.5 g) = mass / molar mass = (2.5 g) / (256.52 g/mol) = 0.009746 mol.

  • We need to determine the limiting reactant,

The theoretical ratio of (Fe: S₈) is (16: 3) which is 5.333.

The actual ratio of the reactants (Fe:S₈) is (0.0537 mol: 0.009746 mol) which is 5.5.

  • This means that Fe is found with a slight increase and S₈ is the limiting reactant.
  • Now, we can calculate the no. of moles of produced Fe₂S₃;

Using cross multiplication:

3.0 moles of S₈ produces → 8.0 moles of Fe₂S₃, from the stichiometry,

0.009746 moles of S₈ produces → ??? moles of Fe₂S₃.

  • The no. of moles of the produced Fe₂S₃ = (8.0 mol)(0.009746 mol) / (3.0 mol) = 0.026 mol.
  • Finally, we can get the mass of the the produced Fe₂S₃.

The mass of the the produced Fe₂S₃ =  no. of moles x molar mass = (0.026 mol)(207.9 g/mol) = 5.405 g ≅ 5.5 g.

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