Determining The Limiting Reactant and Percent Yield LAB

Dear Readers,

I know we haven't always had the best time together, but I'm willing to put a filter on it.

We can work through this together; 

• We'll observe a double replacement reaction
• We'll find the limiting and excess reactant together
• We'll determine the theoretical mass of the precipitate 
• And we'll compare the actual mass with the theoretical mass 
• TOGETHER! WE CAN ACCOMPLISH THE OBJECTIVES OF THIS LAB!!

LET US BEGIN

1. We obtain Na2Co3 solution and CaCl2 solution
2. Mix the two together in a beaker and observe! 
3. Set up a ring stand and funnel with folded filter paper
4. We now swirl the beaker and slowly pour some into the paper funnel. We'll take this slowly,     allowing the solution to smoothly filter out.
5. When we're finished filtering, remove the filter paper and allow it to dry.
6. We will weigh the paper next time.
7. With the power of grayskull (and the filter paper)we can find our excess and limiting reactant!

This is what we saw during the lab

1 Na2CO3(aq) + 1 CaCl2(aq) --> 2 NaCl(aq) + 1 CaCO3(s)

This is how we will find our percent yield

Percent Yield=   actual mass produced (grams)    x 100

           theoretical mass produced (grams)

AND that my readers, was lab 6D. Are we alright now? We can still be friends? I'm horrible with rejection. I'll see you next time..hopefully. I HAVE COOKIES.

Excess and Limiting Quantities

The chemical equation for a reaction describes what is supposed to occur. However sometimes there is not enough of one reactant for the full reaction to occur. The reactant is called the limiting reactant because it limits how far the reaction can go. The limiting reactant is always fully used up. The other reactant(s) are called the excess reactant because there are more than enough moles for the reaction to occur.





Now how do we find out which reactant is which?
There are two ways: the first way is to convert both reactant to the same product and see which one produces the least, the second way is to convert one reactant to the other and see how much is needed to react with each other.

If you have 67.0g of Cl₂ and 35.0g of O₂ in the reaction 2Cl₂ + O₂ à2OCl₂ which reactant is the limiting quantity?

67.0g Cl₂ x 1 Mol Cl₂ x 2 Mol OCl₂ x       87.0g        = 82.1 g of OCl₂

                     71.0g         2 Mol CL₂        1 Mol OCl₂

44.0g O₂ x 1 Mol O₂ x 2 Mol OCl₂ x       87.0g        = 239 g of OCl₂

                     32.0g         1 Mol O₂        1 Mol OCl₂

 This means the Cl2 is the limiting reactant because it can only make 82.1g of OCl2

Now the second way using the same question

44.0g O₂ x 1 Mol O₂ x 2 Mol Cl₂ x       71.0g        = 195 g of Cl₂

                     32.0g         1 Mol O₂        1 Mol Cl₂

Now we can see that we would need 195g of Cl2 in order for all of the O2 to be used up. But we only have 67.0g and therefore we know that Cl2 is the limiting reactant.

Using the second way we can also determine how much excess we have or how much is missing for the whole reaction to occur. 

195- 67= 128g

This means that 128g of Cl is missing.

Here is another example question and tutorial

http://www.khanacademy.org/video/stoichiometry--limiting-reagent?playlist=Chemistry

STOYK STOYK STOYK STOYK

ATTENTION ALL STOYK LOVERS

there is a quiz next class

be prepared.

PRACTICE QUIZ

PRACTICE QUIZ ANSWERS

HAVE FUN

-flees-

-JY

Stoichiometry with Molarity and STP

The past is coming back to haunt us!

MOLARITY AND STP!

this is insane

Let us review what we know about molarity!

• Molarity = moles/litres
• We can also find
• Litres = moles/Molarity
• mole = Molarity x Litres

How about an example? Your negative response has not been heard!

Consider the following equation: Ca(OH)2 + 2 HCL --> CaCl2 + 2H2O

How many litres of 0.100M HCL would be required to react completely with 5.00g of Calcium Hydroxide?

1) Write a balanced equation (it is already balanced)
2) Create a map that gets you from grams of Ca(OH)2 to litres of HCL 
3) 5.00g Ca(OH)2  x  1 mol Ca(OH)2  x   2 mol HCL  =  0.0135 moles HCL

                                     74.1g Ca(OH)2   1 mol Ca(OH)2  

4) Now use the molarity equation to find the amount of litres required

            Litres = 0.0135 moles HCL

                                       0.100M

             Litres = 0.135 L of 0.100M HCL        

Now how about the STP?

Consider the following equation: 2 H2 + O2 --> 2 H2O

Calculate the volume of oxygen gas required to burn 250.0L of Hydrogen gas at STP.

250.0L H2  x  1 mol  x  1 mol O2  x  22.4L   = 125.0L O2

                        22.4L      2 mol H2  1 mol O2

Here is a game to aid you on your quest to conquer the past!

For moles and stoichiometry!

Stoichiometry Part II

CHARLIE SHEEN WELCOMES YOU WITH A WITTY REMARK.

-insert funny remark countering greeting remark-

Stoichiometry

Pronounce stoichiometry as “stoy-kee-ah-met-tree,” if you want to sound like you know what you are talking about, or “stoyk:,” if you want to sound like a real geek.

Stoichiometry calculations involve particles, moles and mass. Yes, all readers. The lovely mole on your nose is back.

NOW LET US BEGIN OUR JOURNEY INTO STOYK:

Write a balanced equation

• you will need it for your mole ratio
• if you do not have a balanced equation you will be WRONG, and you don't want to be wrong

Create a 'road map'

• stating where you are starting and where you want to go
• many stoyk problems will follow this pattern:

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

• you can start anywhere along this map

Do the Calculations

• time to figure out the solution to your problem!

I believe an example is in order!

BAM, example presented:

I believe a different method of learning is in order as well!

BAM! interactive tutorial presented:

with lots and lots of clicking!

Are you feeling the stoyk? No? Re-read this post! and don't come back until you feel geeky enough to say stoyk.

JY

Stoichiometry

Stiochio --> element
Metry----> measurment

Therefore stoichiometry is the theoretical measurement of elements.
Now what is that supposed to mean?
Well, we can use stoichiometry to determine the amount of product a reaction will produce and vice versa.

First of all we need to look at the chemical equation we are using.
Eg. NH₃ + O₂ à H₂O + NO

Which now needs to be balanced correctly
Eg. 4NH₃ + 5O₂ à 6H₂O + 4NO

The coefficients serve as the mole ratio.
This means that when there are 4 moles of NH₃ and 5 moles of NO they will 6 moles of water and 4 moles of NO.

Now the fun part starts
Generally you will be given some amount of one of the reactants/products and have to figure out how much of  another molecule it will produce.
Using the mole ratios you multiply the amount you have by a fraction with what you are trying to find out on the top and the amount you already have on the bottom. This will give you the number of moles you are trying to figure out.

Eg. If you have 4.3 moles of NH₃ how many moles of H₂O will be produced?

4.3 Mol NH₃ x 6 Mol H₂O   = 6.4 Mol of H₂O
                        4 Mol NH₃
 Here is another example


This website is an excellent resource with tutorials and videos
http://www.chemcollective.org/stoich/reaction_stoi.php

Another example is provided in this video
http://www.khanacademy.org/video/stoichiometry-example-problem-1?playlist=Chemistry