Question

Zinc Sulfide reacts with oxygen according to the reaction:

2ZnS (s) + 3 O2(g) 2 ZnO (s) +2 SO2 (g)

A reaction mixture contains 4.2 moles of zinc sulfide and 6.8 moles of oxygen. Once the reaction occurred as a completely as possible, what amount in moles is left of the excess reactant?

Answer 1

The amount in moles of the excess reactant left is 0.5 mole

Balanced equation

2ZnS (s) + 3O₂(g) --> 2ZnO (s) + 2SO₂(g)

From the balanced equation,

2 moles of ZnS reacted with 3 moles of O₂

How to determine the excess reactant

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.

How to determine the mole of the excess reactant remaining

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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