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.

explanation:

venn diagram is pictorial representation with of circles showing the relation between two finite sets or things. the intersection part of two circles conveys the relation present in between those sets or things.

here in the venn diagram of sand and salt two circles represent sand and salt. but the area falling under the intersection of these circle represents the mixture of sand and salt.

so, in a venn diagram the mixture will be placed in a middle that is in between the intersection of two circles.