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lara [203]
4 years ago
12

2. Consider the following reaction: 4 FeS2 + 11O2 → 2 Fe2O3 + 8 SO2

Chemistry
1 answer:
belka [17]4 years ago
6 0
<h3>Answer:</h3>

Oxygen gas (O₂) is the rate limiting reactant and FeS₂ is the excess reactant.

<h3>Explanation:</h3>

From the questions we are given;

4FeS₂(s) + 11O₂(g) → 2Fe₂O₃(s) + 8SO₂(s)

  • Moles of FeS₂ are 26.62 moles
  • Moles of Oxygen, O₂ are 59.44 moles

We are supposed to determine the limiting and excess reactants;

  • From the equation of the reaction given; 4 moles of FeS₂ required 11 moles of Oxygen gas.

Working with the amount of reactants given;

  • 26.62 moles of FeS₂ will require 73.205 moles of O₂ and only 59.44 moles of O₂ are available.
  • On the other hand 59.44 moles of O₂ requires 21.615 moles of  FeS₂, and we are given 26.62 moles of FeS₂ which means FeS₂ is in excess.

Conclusion;

We can conclude that Oxygen gas (O₂) is the rate limiting reactant and FeS₂ is the excess reactant.

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Actually, we can answer the problem even without the first statement. All we have to do is write the reaction for the production of sulfur trioxide.

2 S + 3 O₂ → 2 SO₃

The stoichiometric calculations is as follows:

7 g S * 1 mol/32.06 g S = 0.218 mol S
Moles O₂ needed = 0.218 mol S * 3 mol O₂/2 mol S = 0.3275 mol O₂
Since the molar mas of O₂ is 32 g/mol,
Mass of O₂ needed = 0.3275 mol O₂ * 32 g/mol = 10.48 g O₂
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