Question

If 1.72 mol of ZnS is heated in the presence of 3.04 mol pf O2,which is the limiting reactant?

Answer 1

The limiting reactant is the ZnS

Step-by-step explanation:

The equation for this reaction is:

2 ZnS + 3O₂ = 2 ZnO + 2 SO₂

2 moles of zinc sulfure reacts with 3 moles of oxygen.

Then, 1.72 mol of ZnS would react with ( 1.72 .3)/2 = 2.58 moles of O₂

If we have 3.04 moles, then the oxygen is the reactant in excess.

Let's confirm, the ZnS as the limiting reactant.

3 moles of oxygen react with 2 moles of sulfure.

Then, 3.04 moles of O₂ would react with (3.04 .2) / 3 = 2.02 moles of ZnS

We have 1.72 moles of Zn S and it is not enough for the 2.02 moles that we need, for the reaction.