COPPER (II) CHLORIDE AND IRON LAB

Today we started our first out of three days for the Copper (II) Chloride and iron lab. Basically today we polished a nail by rubbing it with steel wool, took its mass, took mass of an empty baby food jar, and filled it with copper chloride and water. Then we stuck the polished nail into the baby food jar. Tomorrow we are supposed to see a change on the nail; my guess is that it is start to change from iron to copper because it is set up like a single replacement reaction. At the end we are supposed to figure out whether the iron had a 2+ or 3+ charge by calculating percent yield. I am excited to see what happens to the nail!

*****side note******
The reason why the title is in all caps is because the directions for the labs were written this way. I felt like I was being screamed at when I was reading them, but I thought it was kind of funny.

For more information on the lab we started today, watch the following video to clarify the steps we took! Enjoy! Lab Explanation

Is this real life?

Today we took the Stoichiometry quiz which contained 10 question over what we learned in class the past two days. Personally, I felt very confident about my test because it was the first time I actually understood how to do every problem and I finished with lots of time to spare. I am pretty sure I was the first one done too! Trust me, that will NEVER happen again. This test just made my day great! All in all, I either did really good on the test or really bad because I did every question almost the same way every time. I hope it is the first option.

Hopefully everyone did well on it. Good luck!

Finding the Limiting Factor

Limiting factor- the "guy" that runs out first
Excess factor- the "guy" that remains after the equation has taken place
Guy- Mrs. Frankenbergs name for the chemicals/elements in the reaction

Two approaches for finding limiting factor:


Tips: If you are given the mass of a reactant in moles use approach 1. If it is given in grams, use approach 2.
Work out both reactants as you learned in the last post. At the end figure out which reactant has the least amount of product, this is considered your limiting factor. The other reactant is considered your excess factor.  For example:

In this example, Nitrogen would be the limiting factor because it only made 5.7 g NH3.
The excess factor would be hydrogen because it made the greatest amount of NH3 (55g).

Some problems may ask you to find the amount of left over excess material. Here is an example below (number 4 is like previous problems and number 6 is the new type of question that refers to number 4):

If you need more help click on the links below:
Finding the Limiting factor 
Finding the amount left over

Stoichiometry

Overall this Unit has not been that bad. I agree with Mrs. Frankenberg in that it does get a bad rap for how easy it actually is.

Stoichiometry is basically just a formula that you follow for every problem.

1. Write balanced chemical equation
2. Use backbone (shown in picture below) there can be steps that come before or follow the backbone steps, however, this process will be included no matter what



made in Word

• take grams of substance A
• use molar mass of A using periodic table
• moles of substance A
• Use coefficients of A and B from balanced equation
• moles of substance B
• use molar mass of B
• Grams of substance B

 I will take you through a problem we did in class: You have been hired by an Aluminum foil company to study the reaction of Aluminum with oxygen in the air. Someone has found a use for the aluminum oxide solid that is formed from the reaction and wants you to explore possibility for producing this compound. Your job is to study the relationship between grams of each reactant and products by solving a series of stoichiometry equations.

Questions that go with Prompt:

1. Write the balanced equation you will use for the following calculations.
2. If you were given 74.00 grams of aluminum, how many grams of your product would be produced?
3. If you had 64.00 grams of oxygen available, how many grams of aluminum would you need to react with it?
4. How many grams of product would be produced in the reaction describes in #3?
5. If  you wanted to produce 4.00 grams of the product how many grams of oxygen would be needed for the reaction?
6. If 6.54 grams of aluminum were used in the reaction, how many grams of oxygen would be needed?

Answers:

1.  4Al(s) + 3O2(g) ------> 2Al2O3
2.  74.00g Al (given) x (1 mol Al/ 26.98g Al) x (2 mol Al2O3/4 mol Al) x (101.96g Al2O3/1 mol Al2O3)= 139.8 g Al2O3
3. 64.00g O2 x (1 mol Al/32.00gO2) x (4 mol Al/3 mol O2) x (26.98g Al/1 mol Al)= 71.95 g Al
4. 64.00g O2 x (1 mol Al/32.00gO2) x (2 mol Al2O3/3 mol O2) x (101.96g Al2O3/1 mol Al2O3= 137.9 Al2O3
5. 4.00g Al2O3 x (1 mol Al2O3/101.96g Al2O3) x (3 mol O2/2 mol Al2O3) x (32.00g O2/1 mol)=1.88g O2
6. 6.54g Al x (1 mol/ 26.98g Al) x (3 mol O2/4 mol Al) x (32.00g O2/1 mole O2)=5.82g O2

For more practice or help understanding this concept, Click on the links below:
Bozeman Science: Stoichiometry video
Explanation of how to solve more problems (website)
Take a Quiz!