How many grams of oxygen are required to reach with 0.125 moles of zinc sulfide?

6 g of O₂

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2ZnS + 3O₂ —> 2ZnO + 2SO₂

From the balanced equation above,

2 moles of ZnS reacted with 3 moles of O₂.

Next, we shall determine the number of mole of O₂ required to react with 0.125 mole of ZnS. This can be obtained as follow:

From the balanced equation above,

2 moles of ZnS reacted with 3 moles of O₂.

Therefore, 0.125 mole of ZnS will react with = (0.125 × 3)/2 = 0.1875 mole of O₂.

Thus, 0.1875 mole of O₂ is needed for the reaction.

Finally, we shall determine the mass of 0.1875 mole of O₂. This can be obtained as follow:

Mole of O₂ = 0.1875 mole

Molar mass of O₂ = 2 × 16 = 32 g/mol

Mass of O₂ =?

Mass = mole × molar mass

Mass of O₂ = 0.1875 × 32

Mass of O₂ = 6 g

Therefore, 6 g of O₂ is required for the reaction.