Zinc sulfide and oxygen gas react to form zinc oxide and sulfur dioxide. Determine the amount (g) of SO2 that should be produced in a reaction between 32.5 g of ZnS and 23.3 g of oxygen gas. What is the mass of the xs reactant?

7.30g O₂ are in excess and 21.3g of SO₂ should be produced.

Explanation:

The reaction of Zinc sulfide (ZnS) with oxygen (O₂) produce ZnO and SO₂ as follows:

2ZnS + 3O₂ → 2ZnO + 2SO₂

Using this reaction we can find the amount of SO₂ produced converting the ZnS and O₂ to moles and finding limiting and excess reactant as follows:

Moles ZnS -Molar mass: 97.47g/mol-:

32.5g * (1mol / 97.47g) = 0.333 moles ZnS

Moles O₂ -Molar mass: 32g/mol-:

23.3g O₂ * (1mol / 32g) = 0.728 moles O₂

For a complete reaction of the 0.333 moles of ZnS are required:

0.333 moles ZnS * (3mol O₂ / 2mol ZnS) = 0.500 moles O₂.

As there are 0.728 moles of O₂, limiting reactant is ZnS and excess reactant is O₂.

The moles and mass of O₂ in excess are:

0.728mol - 0.500mol = 0.228moles O₂ * (32g / mol) = 7.30g O₂ are in excess

The produced mass of SO₂ -Obtained from the moles of ZnS- is:

0.333 moles of ZnS * (2mol SO₂ / 2mol ZnS) = 0.333 moles SO₂

0.333 moles SO₂ * (64.066g / mol) =