What mass of O2 was used up in the reaction with an excess of SO2 gas if 14.2 g of sulfur trioxide is formed?

2.84g

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2SO2 + O2 - > 2SO3

Step 2:

Determination of the mass of O2 that reacted and the mass of SO3 produced from the balanced equation.

This is illustrated below:

Molar mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 1 x 32 = 32g

Molar mass of SO3 = 32 + (16x3) = 80g/mol

Mass of SO3 from the balanced equation = 2 x 80 = 160g

Summary:

From the balanced equation above,

32g of O2 reacted to produce 160g of SO3.

Step 3:

Determination of the mass of O2 needed to produce 14.2g of SO3.

This can be achieved as shown below:

From the balanced equation above,

32g of O2 reacted to produce 160g of SO3.

Therefore, Xg of O2 will react to produce 14.2g of SO3 i. e

Xg of O2 = (32 x 14.2) / 160

Xg of O2 = 2.84g

Therefore, 2.84g of O2 is needed to produce 14.2g of SO3.