# MAT 154 Topic 5 Project Assignment

December 18, 2018

**MAT 154 Topic 5 Project Assignment in $24 Only**

Assignment Overview

This week’s project focuses on using Algebra to explore the deep meaning of equal sign, coefficients, and variables. This purpose is achieved through the application of systems of linear equation to balance chemical reactions. This task is particularly important for all those learners that will have to take Chemistry in their curriculum. However, the ability to choose the right coefficients and equations is pivotal in making correct data driven decisions, independently of your field of study or career.

Balancing a chemical reaction

Chemical reactions are akin to culinary recipes, in the sense that different ingredients (called reactants) are combined to form meals (called products). An example of a chemical reaction is shown below in the text. In this example, we can see the reactants on the left hand side of the arrow (i.e., CO2 + H2O) and the products on the right hand side of the arrow (i.e., C6H12O6 + O2).

Analogously, we can advocate for the need of chemical reaction balancing by comparing it with the following recipe for buttermilk biscuits:

“Mix flour, baking soda, salt, butter, and buttermilk. Put mixture in a pan and bake for 10-12 minutes”.

This recipe is basically useless, since we don’t know the quantity of each ingredient that we need to mix and bake to produce delicious buttermilk biscuits.

This idea applies also for chemical reactions, with fancier rules and harder math than cooking….

The law of conservation of mass states that matter is not created nor destroyed (we can’t mix and bake a few ingredients and hope to get biscuits). Its consequence in chemical reactions is that all the atoms in the reactant need to be accounted for in the product. For example, let us consider one of the most vital reaction in nature:

CO2 + H2O – C6H12O6 + O2

This reaction is the one of photosynthesis where the sunlight leads to the formation of glucose (C6H12O6) and oxygen (O2) from carbon dioxide (CO2) and water (H2O). As stated, this reaction is unbalanced. Why? Because the total reactants consist of 1 atom of Carbon, 3 atoms of Oxygen and 2 atoms of hydrogen. While the total product consists of 6 atoms of carbon, 8 of oxygen and 12 of hydrogen. How do we balance it? There are many ways, from a very intuitive to a very systematic/mathematical way. Foremost, how would a balanced equation look like? There will be a number in front of every reactant and every product representing how much of each compound is needed in the reaction. Something like this:

xCO2 + yH2O – wC6H12O6 + zO2

Where the coefficients x, y, w, and z are numbers representing the number of each compound. Let’s discuss how to find these coefficients.

Intuitive balancing approach

As is often the case with mathematics, its use becomes less practical if a problem is simple enough to be solved through some critical thinking and some intuition. In the photosynthesis reaction there are only 3 elements (C, O and H) and 4 compounds (CO2, H2O, C6H12O6 and O2), so it is fair to expect that we can balance this equation through some trial and error.

We could start by looking at elements that have only one instance on the left hand side and one instance on the right hand side, we have two options C(Carbon) or H (Hydrogen). There is one atom of C on the left and 6 atoms of C on the right, so it seems that we need 6 CO2 to balance C. There are 2 atoms of H on the left and 12 on the right so it seems that we need 6H2O to balance H.

This is how the equation looks like after doing this analysis:

6CO2 + 6H2O – C6H12O6 + O2

The carbon and hydrogen are balanced, but not the oxygen (O). Should we start from scratch? Of course not, we just look at what can be done to balance O. In particular, after adding the new coefficients, we have 18 atoms of O on the left and only 8 on the right. How do we get from 8 to 18? We need to notice that we cannot touch the atoms in glucose (C6H12O6) since that would unbalance C and H. We need 18 atoms of O, 6 will come from the glucose (C6H12O6), the remaining 12 from O2. To get these 12 atoms of oxygen we will need to add a coefficient of 6 in front of O2 and the balanced equation becomes:

6CO2 + 6H2O – C6H12O6 + 6O2

Let’s now look how to get the same result through a more systematic approach.

Systematic approach to Balancing

We can reach the same result above by creating a system of linear equations whose solution will give the coefficients that balance the reaction. In particular, the system of equations will have as many equations as elements in the reaction and as many variables as the number of compounds in the reaction. The photosynthesis reaction has 3 elements in it (C, O, and H) and four compounds (CO2, H2O, C6H12O6 and O2), so the system will have 3 equations and 4 variables. The variables are the coefficients that will end up in front of the compounds:

xCO2 + yH2O – wC6H12O6 + zO2

and the equations will come from the number of atoms involved in the reaction for each element. Let’s start building them, beginning with Carbon: there is only one in atom of C (in one compound) on the left hand side, so the left hand side of the equation for the carbon will be:

x=….

Since x is the coefficient in front of the compound with C. Similarly, there are 6 atoms of C in one compound on the right hand side of the reaction, so the full equation for C becomes:

x=6w

With similar reasoning we find that the equation for H is:

2y=12w

While the oxygen equation is a bit trickier since it involves 4 different compounds:

2x+y=6w+2z

In conclusion, this is the system that we need to solve:

x=6w

2y=12w

2x+y=6w+2z

There are infinite solutions to this system (why?), and many methods to find them. How do we choose the correct one?

First we need to realize that x,y,w and z need to be positive integer. Then we choose a value that leads to the smallest value of x,y,w and z. This value is obviously 1 and it should be w=1 since x and y depend solely on w. By replacing w with 1 we get:

x=6

2y=12

2x+y=6+2z

Solving leads to x=6, y=6 and z=6. The same solution as the intuitive approach.

Here is what this will look like in Excel (Video for matrix operations in Excel)

A

# X

B

Example from assignment

1

0

0

x

6

x = 6

0

2

0

# y

12

2y = 12

2

1

-2

z

6

2x + y – 2z = 6

A–1

# B

X

1

0

0

6

6

0

1

1

# 12

6

1

0

-1

6

6

Now let’s get to the Project 5 Assignment.

Analyze the Solutions(s) of the System

Use Table 1 to select the chemical reaction based off the last digit of your student ID then perform the following:

• Represent the balancing problem as a system of linear equations.

• From a mathematical perspective, how many solutions do you expect the system to have?

• How do you choose a solution?

• Set up the system in matrix form to be solved in MS Excel (see the many resources on internet on solving a system of linear equations using MS Excel) and solve the system using MS Excel.

• Write the balanced chemical reaction.

• Try and balance the reaction using intuition rather than a mathematical procedure.

• Discuss the benefits and limits of the two approaches (algebraic and intuitive) when balancing equations.

Last Digit of Student Id

Chemical Reaction

0 Li3N + NH4NO3 → LiNO3 + (NH4)3N

1 C5H12 + O2 → CO2 + H2O

2 KOH + H2SO4 → K2SO4 + H2O

3 NH3 + H2SO4 → (NH4)2SO4

4 FeS2 + Cl2 → FeCl3 + S2Cl2

5 MgCl2 + Li2CO3 → MgCO3 + LiCl

6 C4H6O3 + H2O → C2H4O2

7 Na3PO4 + KOH → NaOH + K3PO4

8 C7H16 + O2 → CO2 + H2O

9 NaBr + Ca(OH)2 → CaBr2 + NaOH

Table 1: choose appropriate reaction for Project 5 based off last digit of student ID.

While GCU format is not required for the this assignment, solid academic writing is expected, and documentation of sources should be presented using GCU formatting guidelines, which can be found in the GCU Style Guide, located in the Student Success Center.

This assignment uses a rubric. Please review the rubric prior to beginning the assignment to become familiar with the expectations for successful completion.

You are not required to submit this assignment to Turnitin.

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