March 15th

Guess what guys? There are no stoichiometry calculations today! Instead, we are doing a lab!

A lab? I can hardly wait!

Lab 6D:

Determining the Limiting Reactant and Percent Yield in a Precipitation Reaction

In this lab, our main objectives are to observe the double replacement reaction of sodium carbonate and calcium chloride, to determine which reactant is the limiting one and which reactant is the excess one, and to obtain a theoretical mass and an actual mass, and then calculate the percent yield using the two pieces of information.

Phew! That was a mouthful!

In this lab, we need two 25mL graduated cylinders...

...and one 250mL beaker.

We'll fill one of the cylinders with a Na2CO3 solution, and the other one with a CaCl2 solution. Then we pour both of these solutions into the beaker and wait for five minutes while the solution sits.

Next, we take a piece of filter paper and put our group names on it with a pencil. We weigh the filter paper, then fold it into a fan and place it into a filter funnel. Finally, we pour the solution into the filter funnel, and then remove the wet filter paper filled with the CaCO3 precipitate. We put that on a paper towel in a safe place and wait for it to dry.

A new formula is introduced in this lab to calculate percent yield:

% Yield:

actual mass produced (in grams)           x     100%

theoretical mass produced (in grams)        

Now that we have our formula, we need our mole ratio. How do we obtain that? Simple! Using the word equation, we can write out a double replacement equation!

1 CaCl2(aq) + 1 Na2CO3(aq) ----> 

1 CaCO3(s) + 2 NaCl(aq)

 

 

Next time, we'll measure out the CaCO3 we have left, and we'll use the % yield formula to calculate the theoretical yield!



March 11, 2011

Why hello, hello, hello. Today I am just going to get straight to it because first of all it is Monday today which just says it all and I am tired. But don't worry, I still gaurentee you that I will not dissapoint any of you fellow bloggers and viewers.

Excess and Limiting Reactants
* Excess Quantity
- a balanced equation that describes what should happen in a chemical reaction.
- one reactant  is the excess quantity and some of it will be left over. The second reactant is used up completely, and is called the limiting reactant.

- Calculating amount of products formed when one of the reactants is in excess.
ex 1. How many grams of OCl2 will be formed when 44.0grams of O2 rect with 97.0grams of Cl2
= convert both reactants to the desired product and the smaller amount of product will actaully be produced
= 1O2 + 2Cl2 (arrow) 2OCl2
= convert Cl2 to OCl2 = 97.0g Cl2 X 1 mol Cl2  X 2 mol OCl2 X 87.0g OCl2
                                                      71.0g Cl2     2 mol Cl2      1 mol OCl2
= 118.86grams OCl2
= convert O2 to OCl2 = 44.0gO2 X 1mol O2 X 2 mol OCl2 X 87.0gOCl2
                                                    32.0g O2   1 mol O2       1 mol OCl2
= 239.25g OCl2
= Cl2 is the limiting quantity and O2 is the excess quantity

- Finding limiting quantity, excess quality, and amount excess.
ex 2. When 41.0g of O2 is reacted with 164g of Cl2, how many grams of which rectant will be left over (in excess)
= 1O2 + 2Cl2 (arrow) 2OCl2
= calculate how many grams of Cl2 would be required to react with 41.0g of oxygen gas
= 41.0g O2 X 1 mol O2 X 2 mol Cl2 X 71.0g Cl2
                    32.0g O2    1 mol O2     1 mol Cl2
= 181.93g Cl2
= since there are only 164g of Cl2, not all 41.0g of O2 can react therefore, chlorine is the limiting and oxygen is the excess.
= however, to calculate how much of the excess quantity is left over, we need to find out how much of O2 would react with the limiting reactant (164g Cl2)
= convert Cl2 to O2 = 164g Cl2 X 1 mol Cl2 X 1 mol O2 X 32.0gO2
                                                  71.0g Cl2    2 mol Cl2   1 mol O2
= 36.958g O2
= 37.0g O2 would react with 164g Cl2
= 41.0g (have) MINUS 37.0g O2 (reacts)
= 4.0g O2 would be excess when 164g Cl2 reacts with 41.0g O2.

Whewww man that was alot of calculations!! But dont stress out if you dont get this on the first try; it is perfectly normal and trust me, practice makes perfect!!!
Here are some more excercises that you can & should check out:
1) http://www.science.uwaterloo.ca/~cchieh/cact/c120/limitn.html
2) http://www.science.uwaterloo.ca/~cchieh/cact/c120/stoichio.html
3) http://www.ausetute.com.au/exceslim.html
4) http://wiki.answers.com/Q/How_do_you_solve_limiting_reagent_problems
5) http://members.shaw.ca/mrkwan/chemistry11/Excess%20Reactant%20Notes.pdf

Here are some videos if you are more of a visual learner:
1) http://www.youtube.com/watch?v=JOQz2rIFpVM
2) http://www.youtube.com/watch?v=rESzyhPOJ7I&feature=fvwrel
3) http://www.youtube.com/watch?v=4OeKFsVRmfQ&feature=related
4) http://www.youtube.com/watch?v=dAkbcRgecBo&feature=related
5) http://www.youtube.com/watch?v=SjQG3rKSZUQ&feature=relmfu
6) http://www1.teachertube.com/viewVideo.php?ref=kvanderveen&title=Limiting_Reactant&video_id=75396

Images:






And of course there is always enough time to have some laughs!

March 7th

Today we'll be doing two NEW but OLD types of stoichiometry calculations! They're OLD because they're done using the concepts we learned before, but there's NEW stuff to be learned as well; we're just introducing stoichiometry to them.

So you can think of these concepts as VINTAGE (get it?) :)

Vintage clothing: always great for a night out with friends!

First, let's do stoichiometry calculations with

MOLARITY

Now, as you may recall, the formula for calculating molarity is M = mol / L, where M is molarity, mol is the number of moles, and L is the volume.

This formula could also be manipulated into mol = M x L or L = mol / M.

Question: Methane gas reacts with oxygen to form carbon dioxide and water. If 0.45 mols of methane reacted to form 1200 mL of water, what is the molarity of the water?

To do this problem, go along the following steps.

1. Write out the word equation.

C

H4

+

O2

--->   C

O2

+

H2

O

2. Balance it!

1C

H4

+

2O2

--->

1C

O2

+

2H2

O

3. Now that you have your mole ratio, know what you HAVE and where you want to GET TO. You HAVE the VOLUME (1200 mL) of the water, you want to calculate the MOLES of the water in order to find the MOLARITY of the water.

0.45 mol CH₄ x 2 mol H₂O/ 1 mol CH₄

= 0.9 mol H₂O with sig figs = 0.90 mol H₂O

4. Now use the molarity formula to figure out the molarity of the water. Since the volume given to you is in mL, divide by 1000 to get a litre quantity.

M = 0.90 mol H₂O / 1.2 L H₂O

= 0.75 M H₂O

And there you have it! Calculating stoichiometry with MOLARITY!

Now let's do stoichiometry calculations with

STP

Now as you may recall, the conversion factor for moles to STP is 1 mol / 22.4 L.

Question: Beryllium fluoride reacts with chlorine gas to form beryllium chloride and fluorine gas. If 8.20L of fluoride gas is produced, how many grams of beryllium fluoride was needed?

To do this problem, go along the following steps.

1. Write out the equation.

BeF₂ + Cl₂ ---> BeCl₂ + F₂

2. Balance it to obtain the mole ratios.

1BeF₂ + 1Cl₂ ---> 1BeCl₂ + 1F₂

3. Now, using the conversion factor for STP, figure out the steps used to get where you want to go.

8.20L F₂   x 1 mole F₂/ 22.4L F₂  x 1 mol BeF₂  / 1 mol F₂  x 47g BeF₂ / 1 mol BeF₂

= 17.2 g BeF₂  (watch your sig figs!)

There! That wasn't so hard, was it? :D

Here's a helpful little PP presentation that will clear up any confusion about stoichiometry conversions using mole ratios, STP, and molarity!

Stoichiometry - Mole Calculations

UNTIL NEXT TIME!

March 4th

More complicated STOICHIOMETRY!

Are you serious?!

Okay, fine.

Recall from the previous day's work that it's possible, using mole ratios, to convert mole-mole problems. Now we are going to learn how to convert from moles to grams and moles to particles and all of that good stuff!

LET'S GET STARTED!

Okay, the first thing is to recall the "mole map" we drew while working with previous chapters. Only this time, you can add stoichiometry and mole ratios to the mix.

This map is all nicely laid out for you, but in a real test or quiz, you won't get a map, so it's best to memorize all of this stuff now.

To convert more complicated quantities, you need to follow a few steps.

1. Make sure you have a balanced equation. You'll need one for the mole ratios.
2. Draw a mental mole map in your mind to know a) what quantity you have and b) the steps you need to get to the quantity you want
3. Write out the steps, cancelling out like quantities.
4. Make sure you have your sig figs in order.

Now let's try a few sample problems.

Example One:

Propane gas reacts with oxygen to form water and carbon dioxide. If 7.38g of oxygen is used in the reaction, how many mols of carbon dioxide will be produced?

1. First, write out the equation you get from the word equation.

C₃H₈ + O₂ ---> H₂O + CO₂

2. Next, balance the equation.

C₃H₈ + 5O₂ ---> 4H₂O + 3CO₂

3. Now, draw a mental mole map.

7.38g O₂ x 1 mol O₂/ 32g O₂ x 3 mol CO₂/ 5 mol O₂

4. Reduce the equation.

7.38g O₂ x 1 mol O₂/ 32g O₂ x 3 mol CO₂/ 5 mol O₂

= 0.138375 mols CO₂ with sig figs = 0.138 mols CO₂

Example Two:

Iron metal is dipped into sulfuric acid to produce soluble iron (III) sulfate and hydrogen gas. If 33 grams of iron metal is used in the reaction, how many grams of iron (III) sulfate will be produced?

1. First, write out the equation you get from the word equation.

 H₂SO₄ + Fe ---> Fe₂(SO₄)₃ + H₂

2. Next, balance the equation.

 3H₂SO₄ + 2Fe ---> Fe₂(SO₄)₃ + 3H₂

3. Now, draw a mental mole map.

 33g Fe x 1 mol Fe / 55.8g Fe x 1 mol Fe₂(SO₄)₃/ 2 mol Fe x 399.9g Fe₂(SO₄)₃/ 1 mol Fe₂(SO₄)₃

4. Reduce the equation.

33g Fe x 1 mol Fe / 55.8g Fe x 1 mol Fe₂(SO₄)₃/ 2 mol Fe x 399.9g Fe₂(SO₄)₃/ 1 mol Fe₂(SO₄)₃

= 118.25g Fe₂(SO₄)₃   with sig figs = 120g Fe₂(SO₄)₃

Now, that wasn't so bad, was it? :)

Here's a link to if you want to do some further stoichiometry problems!

Stoichiometry Problems!

And here's a link for the answers:

Stoichiometry Problem Answers

March 2nd

All about STOICHIOMETRY!

Don't let its name daunt you! Stoichiometry is actually very simple, once you understand it!

Basically, stoichiometry refers to the measurement behind the different quantities of the reactants annd the products. You can think of it as the math behind the chemistry!

MATH! IT'S EVERYWHERE!

In stoichiometry, the chemical equations are always in balance. What does this mean? It means we use a term called the mole ratio!

Let's try a few sample problems.

1 Zn (s) + 2 HCl (aq) ---> 1 ZnCl (aq) + 1 H2 (g)

Now, look at the numbers in front of each element. They are 1:2:1:2. That is the mole ratio! This means that ONE molecule of zinc reacts with TWO molecules of hydrogen chloride to produce ONE molecule of zinc chloride and ONE molecule of hydrogen gas.

C2H5OH + 3O2 ---> 2CO2 + 3H2O

In this one, the mole ratio is what?

You guessed it! 1:3:2:3

Okay, now we're going to get down to some real work! Using the equation given above, do these problems, then check your solution.

OH AND ONE MORE THING: SIG FIGS IS ALWAYS RELEVANT!!! DON'T FORGET YOUR SIG FIGS!!!

Formula: 1 Zn (s) + 2 HCl (aq) ---> 1 ZnCl (aq) + 1 H2 (g)

Question: If a mad scientist had 0.270 moles of hydrogen chloride, how many moles of hydrogen gas will come out of the reaction?

Answer:

0.270 mol Zn x 1 mol H2 / 2  mol Zn

= 0.135 mol H2

Formula: C2H5OH + 3O2 ---> 2CO2 + 3H2O

Question: If the same mad scientist had 5.2 mols C2H5OH, how many mols of water could he make?

Answer:

5.2 mols C2H5OH x 3 mol H2O / 1 mol C2H5OH

= 15.6 mol H2O, but with sig figs = 16 mol H2O

Got it?

Now, here's a handy dandy website for you to practice some more STOICHIOMETRY! Good luck, my budding scientists!

Stoichiometry in All Its Beauty

Always make sure that the equation is balanced! It might not always be balanced FOR you in a test/quiz situation!