The amount in moles of the excess reactant left is 0.5 mole
<h3>Balanced equation</h3>
2ZnS (s) + 3O₂(g) --> 2ZnO (s) + 2SO₂(g)
From the balanced equation,
2 moles of ZnS reacted with 3 moles of O₂
<h3>How to determine the excess reactant</h3>
From the balanced equation,
2 moles of ZnS reacted with 3 moles of O₂
Therefore,
4.2 moles of ZnS will react with =(4.2 × 3) / 2 = 6.3 moles of O₂
From the calculations made above, we can see that only 6.3 moles of O₂ out of 6.8 moles given, is required to react completely with 4.2 moles of ZnS.
Thus, ZnS is the limiting reactant and O₂ is the excess reactant.
<h3> How to determine the mole of the excess reactant remaining</h3>
The excess reactant is O₂. Thus the mole remaining after the reaction can be obtained as illustrated below:
- Mole of O₂ given = 6.8 moles
- Mole of O₂ that reacted = 6.3 moles
- Mole of O₂ remaining =?
Mole of O₂ remaining = (Mole of O₂ given) - (Mole of O₂ that reacted)
Mole of O₂ remaining = 6.8 - 6.3
Mole of O₂ remaining = 0.5 mole
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