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;
<u><em>Using cross multiplication:</em></u>
3.0 moles of S₈ produces<em><u> </u></em>→ 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.
