Zinc (II) sulfide reacts with oxygen according to the reaction: A reaction mixture initially contains 3.0 mol ZnS and 2.0 mol O2. Once the reaction has occurred as completely as possible, what is excess reactant and what amount (in moles) of the excess reactant is left?

Answer: Zinc (II) sulfide (ZnS) is in excess. there will remain 1.67 moles

Explanation:

Step 1: Data given

Zinc (II) sulfide = ZnS

oxygen = O2

Number of moles ZnS = 3.0 moles

Number of moles O2 = 2.0 moles

Step 2: The balanced equation

2ZnS + 3O2 → 2ZnO + 2SO2

Step 3: Calculate the limiting reactant

For 2 mles zinc (II) sulfide we need 3 moles oxygen to produce 2 moles zinc oxide and 2 moles sulfur dioxide

O2 is the limiting reactant. There willreact 2.0 moles.

ZnS is in excess. There will react 2/3*2.0 = 1.33 moles

There will remain 3.0 - 1.33 = 1.67 moles ZnS

Step 4: Calculate products

For 2 mles zinc (II) sulfide we need 3 moles oxygen to produce 2 moles zinc oxide and 2 moles sulfur dioxide

For 2.0 moles O2 we'll have 1.33 moles ZnO and 1.33 moles SO2

Answer: Zinc (II) sulfide (ZnS) is in excess. there will remain 1.67 moles

Excess reagent is the ZnS. After the reaction is complete, 1.67 moles of sulfide remain.

Explanation:

This is the reaction to work with:

2ZnS + 3O₂ → 2ZnO + 2SO₂

Limiting reagent is the oxygen. We confirm it by a rule of three

2 moles of sulfide can react to 3 moles of O₂

Therefore 3 moles of ZnS will react to (3. 3) / 2 = 4.5 moles (we need 4.5 moles of O₂ and we only have 2 moles, that's why the O₂ is the limiting)

Excess reagent is the zinc (II) sulfide

3 moles of oxygen react to 2 moles of ZnS

2 moles of O₂ will react to (2. 2) / 3 = 1.33 moles of ZnS (it is ok, be cause we have 3 moles, and we need only 1.33)

After the reaction is complete (3 - 1.33) = 1.67 moles of sulfide remain.