A Stoichiometry Lab! :D

Today's lab: Lab 6D- Determining the Limiting Reactant and Percent Yield in a Precipitation Reaction.

Objectives? To observe the double replacement reaction, determine which is the limiting reactant and which is excess, determine the theoretical mass of precipitate & compare the actual mass with the theoretical mass and calculate percent yield!

Safety first! Put on your safety goggles/glasses!

Obtain the sodium carbonate solution and calcium chloride solution. Pour the contents into a large graduated cylinder and observe! Let it sit for about 5 minutes. While waiting...

Set up the ring stand with the funnel. Get the filter paper, write your name on it, and weigh it. Then fold the filter paper so it fits inside the funnel and place it inside.(Don't know how to fold the filter paper? Just follow the picture.) Place a graduated cylinder underneath the funnel. Now, swirl the beaker with the contents and now slowly pour some of the solution into the funnel. Do it in stages. It will take some time...tick tock tick tock. After filtering is finished, remove the filter paper and place it on a folded paper towel and let it dry. The next day we will weigh and record the mass of the dry filter paper.

So what happened in this lab? A double replacement reaction occurred! 

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

With this information, you can use stoichiometry to determine which is limiting and which is excess. Then you can predict how much precipitate was formed.

To calculate the percent yield next class, this formula will be used:

Percent Yield=   actual mass produced (grams)    x 100

           theoretical mass produced (grams

Excess and Limiting Reactant Percent Yield

This is going to be a difficult chapter so let's break it down. Today's we'll just learn the excess and limiting reactant and the next day would be percent yield.


Excess Quantity: With a balanced equation, we can tell what will happen in the chemical reaction. Except in some cases some reactants may need more or less of one reactant depending the chemical equation if they don't come together.

Excess Quantity in Chemical Reactions: Calculate the amount of products that are being formed and one              reactant would be an excess and one would be a limiting.

Do you get the picture?

No? It's fine!

First let's look at an analogy:

You're working for McDonald's. 

You have 10 hamburger buns and 5 hamburger patties. 

How many hamburgers can you make?

You'll be able to make 5 hamburgers but what was left over?

Solving the puzzle:

There would be 5 hamburger buns in 5 hamburger patty.

Therefore, the beef patty is known as the limiting ingredient since it limits how many burgers can be made.

Got it? Okay, now let's try some chemistry related examples!

Ex. I20.0 g of hydrogen gas react with 100.0 g of oxygen, which reactants is present in excess and by how many grams?

Step 1: Write a balanced equation.

2H2(g) + 1O2 (g) ---->  2H2O(l) 

Step 2: Find which reactants is excess calculate how many grams of H2 would be required to react with 100.0 g of O2 (g). 

100 g O2 x 1 mol O2  x 2 mol H2 x 2.0 g H2

                   16 g O2      1 mol O2    1 mol H2

= 25.0 grams of H2

Step 3: To calculate how much the excess is leftover we must convert O2 ---> H2 to see how much H2 would need to react with O2.

120.0 g H2 x 1 mol H2 x 1 mol O2 x 16.O g O2

                      2 g H2        2 mol H2     1 mol O2       

= 480 grams of O2     

More examples: 

1.2Al + 6HCl →2AlCl3+ 3H2If 25 g of aluminum was added to 90 g of HCl, what mass of H2will be produced (try this two ways –with a chart & using the shortcut)?

2.N2+ 3H2→2NH3: If you have 20 g of N2and 5.0 g of H2, which is the limiting reagent?

3.What mass of aluminum oxide is formed when 10.0 g of Al is burned in 20.0 g of O2?4.When C3H8burns in oxygen, CO2and H2O are produced. If 15.0 g of C3H8reacts with 60.0 g of O2, how much CO2is produced?

Did you say you wanted to learn more about excess and limiting reactants? Well you've scrolled to the right place! Watch this video to learn more! 

STOICHIOMETRY QUIZ REVIEW!

STOICHIO - ELEMENT
METRY - MEASUREMENT

Stoichiometry deals with "quantitative analysis of chemical reactions and the measurement of elements and compounds involved in a reaction." It is the "study of relationship between the amount of reactants with the amount of products."

Let's analyze this chemical reaction.
We can see that for every 2 moles of Hydrogen gas, 1 mole of Oxygen gas is needed to react. Similarily, for every 2 water molecules produced, 1 mole of Oxygen gas is needed.

These are called MOLE RATIOS, and they become very important in equations.

REMEMBER TO ALWAYS START WITH A BALANCED EQUATION.

Then, put the mole ratio of WHERE YOU ARE GOING over the mole ratio of WHERE YOU ARE COMING FROM.

E.g. If I knew that I had 4 moles of oxygen gas and wanted to find out how many moles of hydrogen I needed, I would multiply 4 by 2 moles of Hydrogen gas/1 mole of oxygen gas.

Here are some other important points to remember.

1) 1 mole = 6.022 x 10^23 molecules/atoms/particles/formula units
2) 1 mole = atom's/molecule's molar mass in grams
3) 1 mole = 22.4L at STP
4) Molarity (M) = moles of solute/volume of solution

YAY STOICHIOMETRY!

Stoichiometry calculations along with Molarity and STP!!

Remember molarity from chapter 4? Well, we're going to apply that to stoichiometry now!

Review: Molarity = mole

                              Litres

So, you can change it to Litres = mole       OR           mole = Molarity x Litres

                                                     Molarity

Lets try an example.

Ex.1 Consider the following reaction: Zn + 2HCl --> ZnCl2 + H2

How many moles of hydrogen gas are needed to completely react with 15 g of zinc?

Step 1:  The reaction is given to you and is already balanced.

Step 2: Your mole map route is going to be from grams of zinc--> moles of zinc --> moles of H2

Step 3: 15g of zinc  x  1 mol zinc  x  1 mol H2  = 0.23 mol of H2 (don't forget sigfigs)

                                     65.4g zinc       1 mole zinc 

Ex. 2 Using the same reaction from above,

What is the molarity of HCl if 50 L of HCl is reacted to form 6.0 moles of H2

Step 1: The reaction is balanced.

Step 2:  Map route will be from moles of H2 --> moles of HCl. Then take the molarity of HCl.

Step 3:  6.0 moles H2  x  2 moles HCl  = 12 moles HCl

                                           1 mole H2

Now that you have moles of HCl, you can find out the molarity by putting moles over litres.

Molarity = 12 moles HCl

                    50 L HCl

              = 0.24 M HCl

Now, the last thing is to calculate stoichiometry along with STP.

Ex. 3 With the following reaction: N2 + 3H2 --> 2NH3

Calculate the number of grams of nitrogen gas required to make 1.44L of ammonia at STP.

1.44 L ammonia  x  1 mole ammonia  x  1 mole N2  x  14.0 g N2  =  0.450 g of N2

                                         22.4 L              2 moles NH3   1 mole N2

Here is a video that may help you!



Stoichiometry CALCULATIONS!

Stoichiometry calculations involve particles-moles-mass. That's right, we're back to mole conversions. Woohoo!

A little "road" map to help you do the calculations!

Step 1: Write a BALANCED equation. This is important to get the MOLE RATIO.

Step 2: Make a "road map". This will help to figure out where to start and where to go to get to the answer you need.

Step 3: Do the calculations!

Remeber SIGFIGS!!!

Let's try an example! 

This equation is given: 1 C3H8 + 5 O2 --> 3 CO2 + 4 H2O

What mass of C3H8 is needed to produce 100.0g of H2O?

Step 1: It is already balanced so move on to the next step!

Step 2: Road map: Mass of H2O --> Moles H2O --> Use the ratio of 1mol C3H8=4mol H2O --> Convert moles of C3H8 to mass C3H8.

Step 3: 

100.0gH2O x 1 mol H2O x 1 mol C3H8 x 44.0g C3H8 = 61.1g C3H8

                        18.0g H2O   4 mol H2O      1 mol C3H8

As you can see, you can't just convert grams to grams. In order to get there, you have to go through moles. In stoichiometry, many calculations follow this pattern:

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

You can start anywhere along this path.

Let's have some more practice!

Using the same equation as above, what is the mass of CO2 produced by reacting 2.00 mol of O2?

2.00 mol O2 x 3 mol CO2 x 44.0g CO2 = 52.8g CO2

                         5 mol O2       1 mol CO2

1.35 x 10^-6 g of C3H8 is extracted from a gas-bearing rock. How many molecules of CO2 are produced if the gas sample is burned in the presence of excess O2?

1.35 x 10^-6gC3H8 x 1mol C3H8 x 3molCO2 x 6.022 x 10^23 moleculesCO2= 5.54 x 10^16 molecules

                               44.0g C3H8   1mol C3H8                 1 mol CO2

Here's a quiz for some extra practice! :D

A quest to perfecting your stoichiometry calculating skills!

Ready for some.... STOICHIOMETRY!!

You might ask what's a stoichio... what?
It's stoichiometry. To break is up STOICHIO it greek for element and METRY means measurement.

Stoichiometry is about measuring the amounts of element and compound in a reaction and it's also about the study of relationship between the amount of the reactants used in chemical reaction and the products produced in a reaction.

It's confusing at first but let's try some examples!

Ex. 3C2H4 + 5O2 ---> 6CO2 + 4H2O (NOTE: make sure the chemical equations are always balance.)

There are several ways to do this, for example, you can look at the chemical formulas that stands for the number of atoms like 3C2H4 => 2C and 1 H atom or 3C2H4 = 48 g/mols. Other ways could be looking at the coefficient 3 X C2H4 = 6 carbon and 12 hydrogen.

Looking at the same example 3C2H4 + 5O2 ---> 6CO2 + 4H2O
Stoichiometry shows us the ratio 3:5:6:4. This ratio is call the mole ratio.  


3:5:6:4 can be expressed as 3 molecules or C2H4 reacting with 5 molecules of O2 producing 6 molecules of CO2 and 4 molecules of H2O. 


(Reminder: As written above all equations should be balanced because the the equation tells us the ratio of molecules in the chemical reaction.)

Oh! You thought the mole wouldn't come up eh? YOU'RE WRONG. The mole will always be where you are...jks! But as for this chapter mole still exist so don't be too excited :)

INFO THAT MAY HELP YOU:
What you need over what you have or where you're going over where you start.
 What you need       or      Where you're going      
What you have                 Where you start

 Got it? Okay, let's start the fun.

3C2H4 + 5O2 ---> 6CO2 + 4H2O

Ex. How many moles of O2 will form when 6.3 mole of H2O is taken?

6.3 mole of H2O x 5 mole of O2   = 7.9 mol of  O2
                               4 mole of H2O

Now try some examples by clicking on this link. Solutions will also be posted. Have fun and MOLES!
Practice Problem (1)
Practice Problem (2)
Solution:
Solution to Problem (1)

Still need a little more understanding of it? Check these out :D ENJOY!