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)
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