Thursday, January 14, 2021

Stoichiometry Walkthrough

General Chemistry Index

Where are we going with this? This page will give the ability to demonstrate an understanding of the law of conservation of mass through the use of particle diagrams and mathematical models.


Stoichiometry Walkthrough
Let's take another look at this stoichiometry thing… 

The essence here is that we are dealing with the quantities of stuff that goes into a chemical reaction. How much of one thing reacts with another thing and how much of what is produced?

In a different article, steps were developed around the quantities being measured in grams. In this one, we'll build on that and go a little more into details.

Let's get going…

Since we are dealing with stuff in a chemical reaction, the first part of the process is to figure out the reaction. 

STEP 1: Begin with a balanced chemical reaction.
Which isn't always that easy, right? All the predicting products and stuff! Finding coefficients… All that!


STEP 2: From the balanced chemical reaction, determine the mole ratio of the reactants and products.

Those coefficients again! And all that math…  

Let's look at a simple reaction as an example (adding the "1" coefficient that chemistry grammar says we don't actually write/type):

2Na + 1Cl2 --> 2NaCl

So, based on that balanced equation, we know that the reactants will combine in a set, fixed molar ratio of 2 to 1 and produce 2 moles of product.

Lost to some students is that the 2:1-->2 ratio is ALWAYS present in the balanced equation. However, it can be any fraction or multiple, so long as the ENTIRE ratio changes. So, all of the following ratios of moles are valid for that reaction.

1 X — 2 : 1 --> 2

10 X —20 : 10 --> 20

.1 X — .2 : .1 --> .2

1/2 X — 1 : .5 --> 1

7 X — 14 : 7 --> 14

.134 X — .268 : .134 --> .268

There are literally infinite sets of numbers that will fit the 2 : 1 --> 2 pattern. Just multiply the whole ratio by any number and the results will fit the pattern.

Once you have the mole ratio from the balanced equation, you are ready to use that with the actual amounts present, needed, or produced.


STEP 3: Find out how many moles of one thing you are working with.

The number of moles (n) is found several ways.

If you are given the mass of something, moles can be found (where n is moles) using…

 n = mass in grams / molar mass


If dealing with gases, then where P is pressure, V is volume, R is a constant, and T is temperature Kelvin, then the ideal gas law can be used:

PV = nRT

Many times, the ideal gas law is estimated to be:

n = V/22.4 liters


If you are working with a solution, then you can use the molarity (M) of the solution to find moles (n). Where V is volume in liters then

M = n/v

M • V = n

so…

n = M • V


Whatever is given, by some means, you figure out how many moles are present of something in the reaction.



STEP 4: Use the number of moles present with the molar ratio to find out how many moles of everything else is needed or will be produced.

Because reactions occur in fixed ratios of moles, and because one mole of any substance has a set mass, it is pretty easy easy to figure everything out.


______________________

EXAMPLE

Starting with 2.3 grams of Na, how much Cl2 will be needed to completely react without any excess?


STEP 1: Balanced Equation

2Na + 1Cl2 --> 2NaCl

STEP 2: Ratio (molar) of coefficients

2 : 1 --> 2


STEP 3: Actual moles present

 Let's use these atomic weights:

Na = 23.0

Cl = 35.5

Convert the grams of Na present to moles present:

moles = grams / molar mass

n = 2.3 / 23

n = .1 moles 

 

STEP 4 : Needed and produced moles

In this step, you are first thinking this: How many moles do I have as a percentage of how many moles I was supposed to have?

So, divide what is present by the amount of that thing needed in the ratio given (STEP 2) by the coefficients of the balanced equation.

decimal amount given  = given moles / needed needed

decimal amount given = .1 / 2

decimal amount given = .05


Then you are thinking this: If I start with (some decimal amount given) of the one thing, then I need that same decimal amount of everything else.

So, multiply the whole molar ratio (STEP 2) by the decimal amount given .


2 : 1 --> 2

      X  .05 

________

.1 : .05 --> .1


Hence, starting with  .1 moles of Na, you need  .05 moles of Cl2 and you will produce .1 moles of NaCl.

It is easy to convert further into grams/liters of Cl2 and/or NaCl.


grams = moles • molar mass

grams Cl2 = .05 X 71          <-- Each Cl has a mass of 35.5 and Cl2 has 2 of them so…

grams Cl2 = .05 X 3.55


Finding grams of NaCl will be left as an exercise for the reader.*


n = V/22.4 liters

.05 = V/22.4 liters

.05 • 22.4 litres = V

1.12 liters = V


SUMMARY:

Stoichiometry is much harder to spell than it is to do. Using the four steps provided, it really is just a matter of multiplying and dividing.

Begin with a Balanced equation

Find the Ratio from the coefficients

Calculate the Actual number of moles present

Calculate the Needed moles for the other reactants and products.


BRAN is good, right?

And, just what does this look like on paper?




__________________

* This footnote has nothing to do with this topic. But…

When in college a zillion years ago, I had this text book that would go through some vastly complex calculation, then skip like 50 steps that were replaced by something like:

Converting from the above relationship to the following is intuitively obvious to the most casual observer.

Nah… never.

Then, it would end up and add something that amounted to this:

Creating ongoing world peace will be left as an exercise to the reader.


No comments:

Post a Comment