Question

Manufacturing is able to produce large quantities of sulfur dioxide, SO2, by heating zinc sulfide in the presence of oxygen yielding zinc oxide and sulfure dioxide.

1. Write a chemical equation.

2. If 1.72 mol of ZnS is heated in the presence of 45.6 g of O2, which is the limiting reactant?

3. How many grams of excess reactant remain?

4. If the typical yield of sulfur dioxide is 86.78%, what would be the actual yield if 4897 g ZnS are used?

Answer 1

1. 2ZnS + 3O₂ ------> 2ZnO + 2SO₂

2. Moles = Given mass/Molar mass

Moles of O₂ = 45.6/32 = 1.425

Moles of ZnS = 1.72 (given)

According to stiochiometric coefficient of reaction 2 mole of ZnS react with 3 mole of O₂.

So number of moles of ZnS required to react with 1.425 mole of O₂ = (2/3)*1.425 =0.95

So from 1.72mole of ZnS 0.95 mole will react and 1.425 O₂ will react completely.

The reactant which reacts completely in the reaction is limiting reagent ,so O₂ is limiting reagent.

3. Moles of ZnS left unreacted = 1.72 - 0.95 = 0.77

Mass of ZnS left unreacted = Moles left unreacted * Molar mass

Mass of ZnS left unreacted = 0.77*97.474 =75.054g

4. Mass of ZnS used for production of SO₂ =4897g

Moles of ZnS = 4897/97.474 = 50.24

Since 2moles of ZnS yield 2 moles of SO₂ if yield is 100%.

If yield is 100% 50.24 moles of ZnS will yield 50.24 moles of SO₂.

But yield is 86.78%, therefore no. of moles of SO₂ formed = (50.24*86.78)/100 =43.6

Mass of SO₂ produced = moles produced*molar mass of SO₂

= 43.6*64 =2790.4g