Limiting and Excess Reagent Problems

Category: Chemical Engineering Math

"Published in Newark, California, USA"

What is the maximum weight of SO₃ that could be made from 25.0 grams of SO₂ and 6.0 grams of O₂ by the following reaction? (Atomic Weights: S = 32 and O = 16)
 

Solution:

The first thing that we need to do is to get the molecular weights of SO₂, O₂, and SO₃ from the given atomic weights.

Molecular Weight of SO₂:

Molecular Weight of O₂:

Molecular Weight of SO₃:

Next, we will calculate the amount of SO₃ which is the product of SO₂ and O₂. From the given reaction 

Weight of SO₃ from SO₂:

Weight of SO₃ from O₂:

From the two reactants, O₂ will give us the least amount of product which is SO₃. O₂ is the limiting reagent. On the other hand, SO₂ is the excess reagent. Once O₂ is completely converted into SO₃, there will be a left over or unused amount of SO₂ in the reaction. 

In this problem, we have to choose the limiting reagent to determine the amount of product produced. Therefore, the maximum weight of SO₃ is

Mole Fraction Problems

Category: Chemical Engineering Math

"Published in Newark, California, USA"

Calculate the mole fractions of ethyl alcohol, C₂H₅OH, and water in a solution made by dissolving 9.2 grams of alcohol in 18 grams of H₂O. (Atomic Weights: H = 1, C = 12, and O = 16)

Solution:

There are two substances or components in a solution which are ethyl alcohol (solute) and water (solvent). The first thing that we need to do is to get the molecular weight of each component.

Molecular Weight of Ethyl Alcohol:

 

Molecular Weight of Water:
 

 

 

Next, we can get the moles of each component since we know their molecular weights.

Moles of Ethyl Alcohol:
            

  

    

                                   
Moles of Water:
                          

                 

                                               

                                                                                             
Therefore, the mole fraction of ethyl alcohol is

and the mole fraction of water is

 

 

Note: Mole fractions are unitless and the sum of the mole fractions of components in a solution must be equal to 1.

More Cylinder Problems, 13

Category: Solid Geometry

"Published in Vacaville, California, USA"

The outer protective smokestack of a steamship is streamlined so that it has a uniform oval section parallel to the deck. The area of this oval section is 48 sq. ft. If the length of the stack is 18 ft. and if the axis is raked 76°15' to the horizontal, find the volume enclosed by the stack. (Raked means inclined.)

Solution:

To illustrate the problem, it is better to draw the figure as follows

Photo by Math Principles in Everyday Life

The area of a base which is an oval section is already given in the problem. By using a trigonometric function of a right triangle, the altitude of the smokestack is

Therefore, the volume of a smokestack which is a cylinder is

More Cylinder Problems, 12

Category: Solid Geometry

"Published in Vacaville, California, USA"

A cylinder whose base is a circle is circumscribed about a right prism of altitude 12.6 ft. Find the volume of the cylinder if the base of the prism is an isosceles triangle of sides 3 ft., 3 ft., and 2 ft.

Solution:

To illustrate the problem, it is better to draw the figure as follows

Photo by Math Principles in Everyday Life

Consider the base of a cylinder which is a circle that circumscribed about an isosceles triangle as follows 

Photo by Math Principles in Everyday Life

The point of intersection of the perpendicular line bisectors of the sides of an isosceles triangle is also the center of a circle. The radius of a circle is equal to the distance from the point of intersection of the perpendicular line bisectors of the sides of an isosceles triangle to its vertex. By using trigonometric functions of right triangles, we can solve for the radius of a circle as follows

 

 

  Hence, the radius of a circle is
 

 

 

Therefore, the volume of a circular cylinder is 

But

Hence, the above equation becomes