6D Determining the Limiting Reactant and Percent Yield in a Precipitation Reaction

 Another lab!! Remember, safety first!! Always put on safety equipment first!!
Today, we conducted an experiment in by mixing two solutions together in order to form a precipitate. Our objectives were to determine which reactant is the limiting reactant and which is the excess reactant, determine the theoretical mass of the precipitate, and to calculate the percent yield.

 First of all, some supplies we needed were the following:

A Centigram

Filter Paper

So anyways....
We poured 25 mL of the Na2CO3 solution and 25 mL of the CaCl2 solution into the the 250 mL beaker. After obtaining a piece of filter paper, we set up a filtering apparatus! After, we use a wash bottle to wet filter paper in the funnel. Proceeding on with the lab, pour the combined solutions into the filter funnel. Then after ,you wait until the filter process is done. After the filtering, you let the filter paper dry. The process looks something like the picture below

Anyways, on the second day we had to weigh the mass of the dried filter paper in order to come up with conclusions.

By using Stoichiometry we learned a few classes before, we determine which reactant is the excess/ limiting reactant. We also can determine how much precipitate formed by calculating the theoretical mass.

After this we calculate the percent yield. A formula that will help get you the percent yield is the following formula:

Percent yield =

actual mass produced (g)              x 100
theoretical mass produced (g)



Our results was that the CaCl2 is the limiting reactant, the excess reactant is the Na2CO3. The theoretical mass is 1.25125 CaCO3 and the percent yield is 85.5% .

Although there is no specific video about this lab, however if you do not understand how to do the calculations, the following videos will be able to help.

Excess and Limiting Reactants

Excess and Limiting Reactants...................





This is possibly the hardest section of this chapter.
So, before we move into the actual calculations, we should know what they are.


Up until this point, we realize balanced equations tell us what happen in the reaction. Today, we know that balanced equations only tell us what should/usually happen in the reaction, but sometimes, the necessary conditions are not present.


For example, maybe the pressure/temperature is different, or the molar concentration is not as great. 
Therefore, it is necessary to add more of one reactant than the equation predicts, because it is not possible for every atom or molecule to come together.


In other words, one reactant will be in excess (some of it will be left over) and another reactant will be used up completely. 


This will be the limiting reactant, because it limits how much of a product we can actually produce.


If a bike is made from 2 wheels and 1 seat, and we have 4 wheels and 5 seats, what is the number of bikes produced?




The answer: 2 bikes...because if we just look at seats, there are 5 bikes possible. However, there are not enough wheels to make 5 bikes. That is why the limiting reactant is the wheel, and the seat is the excess reactant. 


Now...let's get calculating!



We have the reaction:  AgNO3 + CaCl2 --->  AgCl + Ca(NO3)2


When 32.0 g of CaCl2 reacts with 54.2 g of AgNO3, how many grams of AgCl are formed?


First, we balance the equation: 
2 AgNO3 + CaCl2 ---> 2 AgCl + Ca(NO3)2


Now we want to convert both the reactants to the desired product. The smaller amount of product calculated is our answer. 


32.0 g CaCl2 X 1 mole CaCl2 X 2 mol AgCl X   143.4 g AgCl  = 82.6 g AgCl
                              111.1 g             1 mol CaCl2        1 mol AgCl


54.2 g AgNO3 X 1 mole AgNO3 X 2 mol AgCl    X 143.4 g AgCl = 45.7 g AgCl
                             169.9 g           2 mol AgNO3    1 mol AgCl


Clearly, 45.7 g is smaller, so only 45.7 g of AgCl will be produced from this reaction. We can also realize that AgNO3 is the limiting reactant, since there is CaCl2 left over.


Now, if the question was "How much of the excess reactant is left over?" 


We can calculate it. First, we want to see which one is limiting and which one is excess. We have done that above.


Next, since we are looking for how much of the excess reactant is left over, we want to convert the limiting reactant to the excess reactant using stoichiometry calculations.


The answer we get means 54.2 g of AgNO3 equals to ____ g of CaCl2. 


So.....  54.2 g AgNO3 X 1 mole AgNO3 X 1 mole CaCl2       X 111.1 g CaCl2        = 17.7 g CaCl2
                                      169.9 g             2 moles AgNO3    1 mole CaCl2


This means that 54.2 g of AgNO3 is equal to 17.7 g of CaCl2. 


To get the excess, we simply take the excess reactant and minus our answer above. 
32.0 g CaCl2 - 17.7 g CaCl2 = 14.3 g CaCl2


There are 14.3 g of CaCl2 left over, after 32.0 g of CaCl2 reacts with 54.2 g of AgNO3.


Now, it's time for a video and some problems!!!!
http://www.ausetute.com.au/exceslim.html
http://www.chemteam.info/Stoichiometry/WS-limiting-reagent.html
http://www.softschools.com/quizzes/chemistry/stoichiometry_limiting_reactants/quiz1210.html

Stoichiometry Calculations.....NOW WITH MOLARITY AND VOLUME!

OK, so last blog, you learned about doing some simple stoichiometry calculations that involved converting the mass of one substance into the mass of another substance.


Now, let's pull out the mole map


We're going to try some harder conversions: by including Molarity conversions and Molar Volume.

So, as we all remember, molarity is the molar concentration. The formula is M = mol/V



Let's try an example:
We have a chemical equation of
NaCl + CaO ---> Na2O + CaCl2

and we have 120 mL of a 0.300 M solution of NaCl. How many grams of CaCl2 are produced?

First, we balance:
2 NaCl + CaO ---> Na2O + CaCl2

Then, we figure out how many moles of NaCl there are, based on the information given:

M = mol/V   ---> mol = MV
mol NaCl = (0.120)*(0.300) = 0.0360 mol NaCl

0.0360 mol NaCl X 1 mol CaCl2 X 111.1g           =  1.9998 (2 SF = 2.0g CaCl2)
                                 2 mol NaCl     1 mol CaCl2

This answer tells us that 2.0 g of CaCl2 is produced when 120 mL of 0.300 M solution of NaCl reacts with sufficient CaO.

Now, we can also add on Molar Volume calculations.
Recall that there are 22.4L per mole in STP conditions (Standard temperature and pressure)


Let's try another example:
We have the chemical reaction:  Li2O + MgCl2 ---> LiCl + MgO
How many grams of Li2O will produce 9.0 L of MgO at STP?

First, we balance:
Li2O + MgCl2 ---> 2 LiCl + MgO

Now, we convert:
9.0L MgO X 1 mol MgO X 1 mol Li2O X     29.8g      =  11.97 (2 SF = 12 g)
                         22.4 L          1 mol MgO    1 mol Li2O

So this means that 12 g of Li2O reacts with sufficient MgCl2 in order to produce 9.0 L of MgO

Now, here are some problems, as well as a video:

Stoichiometry Calculations!!!

Well.. we're back to moles AGAIN. After this year of chem... I'm gonna go play some whack-a-mole! Not even kidding!

Anyways... Stoichiometry involves particles, mass and MOLES! Well lets start off this blog/lesson with a mole map! Stoichiometry deals with quantitative relationships between a reactant and product. That is why the relations for the quantities can usually produce a ratio of some sort!

Well ok.. so how to do these calculations. Well it is easier to these calculations in steps.

The First Step: Make an equation out of the word problem, and balance it!
The Second Step: You can make a road map to help you visualize what you need to do
The Third Step: Follow the map and calculate what you need to calculate!!!

Well lets learn more about Stoichiometry by doing a few examples first.

Ex 1/ Consider this equation:  2KMnO4 + 16HCl --> 2 KCl + 2MnCl2 + 8H2O + 5Cl2
How many grams of KMnO4 will be needed to react with 2.30 moles of H2O?

 Step 1 is already done for you because it is already balanced!
Step 2:  2.30 moles H2O --> moles KMn04 --> ? grams KMnO4
Step 3:

2.30 moles H2O x 2 mole KMnO4    x  158.0 g KMnO4  = 90.8 grams KMnO4
                              8 mole H2O        

REMEMBER TO DO YOUR SIG FIGS!!!! ANYWAYS! That was easy right???

Well... If you are like me, and is REALLY confused...here is a pattern you can refer to!!!

grams(x)<-->moles(x)<-->moles(y)<-->grams(x)


You can start as well as go anywhere along this path, as long as you do it in the correct order!!
Now that you have this in mind.. lets start do one more example!

Example 2/


Consider this equation: 4KO₂ + 2H₂O --> 4KOH + 3O<sub>2

How many moles of KO2 can be made using  4.50 mole O2?

4.50mole O2 x 4 mole KO2   = 6 mole KO2</sub>                     
                          3 mole O2

K.. You might THINK you had enough of Stoichiometry and moles.. but don't worry it'll come to haunt you next lesson too! Soo.. you might as well do some more practice on these sites!
http://library.thinkquest.org/10429/high/stoich/question.htm
http://www.askmehelpdesk.com/chemistry/stoichiometry-problems-176369.html
http://www.askmehelpdesk.com/chemistry/stoichiometry-questions-113202.html



WELL .. if u don't understand the way I iunno... taught the lesson... well SUCK IT UP! and... i guess you  can watch this video!

Chapter 6 - Stoichiometry

Our next chapter in the intense study of chemistry is about stoichiometry.
Wait, wait, wait.....stoichio wha?????????
The Greek root stoichio means element, and mentry meaning measurement.



So, it's all about element measurements. At least, that's what the word means.
What it is really about is the analysis of chemical reactions, measuring amounts of elements and compounds that are involved in any given reaction. It is the relationship of reactants used vs products created.



E.g. 2H2      +      O2 --->     2H2O     From our previous knowledge, we can say:
    2 g/mol        32 g/mol      18 g/mol
     4 H's              2 O            4 H's 2 O's

From the same example, we can take the coefficients as a number of moles or a number of molecules. So in other words, we can treat them as:
2 moles H2 + 1 moles O2 ---> 2 moles H2O
The ratio we are given, 2:1:2 is called the mole ratio.

In order to perform the stoichiometry calculations, we must have:

A balanced chemical reaction
A ratio of molecules or moles of substances in the chemical reaction (you will get this when the equation is balanced)

The coefficients tells us how many moles of each substance were reacted or produced. This is where the deadly mole conversion factors come back into play!

Let's start with an example:

How about: Na + Cl2 ---> NaCl

So first, let's check if it's balanced.

2Na + Cl2 ---> 2NaCl

Now it is! Ok, so how many moles of NaCl are produced when 12 moles of Na are produced?

We start off writing down 12 moles Na. To get to moles of NaCl, we must multiply by where we want to go over where we started.

In other words: 12 moles Na   X   2 mole NaCl 
                                                      2 mole Na
(There are 2 moles of Na for every 2 moles of NaCl, as shown in the chemical equation.)
We can cancel out moles of Na to get to our answer. Through simple multiplication, we realize that 12 moles of NaCl are created when 12 moles of Na reacts with chlorine.

Now enough of us, and time for you to get cracking!
http://www.sciencebugz.com/chemistry/chprbstoich.html
http://www.standnes.no/chemix/examples/stoichiometry-problems-chemistry.htm