1. Write the balanced chemical equation:

2 CuS + 3 O₂ → 2 CuO + 2 SO₂

2. Convert grams of reactants to moles:

* CuS:

* Molar mass of CuS = 63.5 g/mol (Cu) + 32.1 g/mol (S) = 95.6 g/mol

* Moles of CuS = 100 g / 95.6 g/mol = 1.046 moles

* O₂:

* Molar mass of O₂ = 2 * 16 g/mol = 32 g/mol

* Moles of O₂ = 56 g / 32 g/mol = 1.75 moles

3. Determine the mole ratio of reactants:

The balanced equation shows that 2 moles of CuS react with 3 moles of O₂.

4. Calculate the moles of O₂ needed to react completely with the available CuS:

* Moles of O₂ needed = (1.046 moles CuS) * (3 moles O₂ / 2 moles CuS) = 1.569 moles O₂

5. Compare the calculated O₂ needed with the available O₂:

* We need 1.569 moles of O₂ but only have 1.75 moles. This means we have excess O₂.

Conclusion:

Since we have less O₂ than needed to react with all of the CuS, CuS is the limiting reactant.