1. SKILLS Project Reaction Stoichiometry and Theoretical Yield

2. What is Stoichiometry? Stoichiometry is the study of the relationships between substances in a chemical reaction. We use DA, along with stoichiometry, to find the following: –The amount of products of a chemical reaction (theoretical yield). –The amount of reactants necessary to produce a specific amount of product. –Relationships between moles of different substances.

3. Stoichiometric Conversions Stoichiometry works the same way as normal DA, however, we can use conversion factors from a balanced equation such as: –mol A = mol B –molar mass A = 1 mol A –Any combination of mol/mol conversions along with molar mass conversions.

4. Using Mole/Mole Conversions: Given a balanced chemical equation: 2A + 4B  3C + 5D You can convert moles of any substance in the equation to moles of anything else in that same equation. For Example: 2 mol A = 4 mol B 2 mol A = 3 mol C 2 mol A = 5 mol D

5. Line ‘Em Up! All conversion factors may be flipped and used two different ways. For example: 1 mol NaCl = 58.44 g 50 mol NaCl 50g NaCl 1 mol NaCl 58.44g NaCl mol  g g  mol The same conversion factor is used in both DA’s, but has been flipped to line up with the previous bracket. This reflects the fact that one DA is mol  g while the other is g  mol.

6. Like most all of the problems you will face in this unit, this is a conversion between two substances. We will want to begin with a DA. 3NaOH(s) + H 3 PO 4 (aq)  3H 2 O(l) + Na 3 PO 4 (aq) ___________________________________ Example 1: mol  mol 1 mol H 3 PO 4 (aq) 3 mol NaOH(s) _ _ _ _ 3.26 mol NaOH(s) Given 3.26 mol NaOH(s), calculate the number of moles of H 3 PO 4 (aq) that will be consumed. As you learned in the “Basic DA” unit, we’ll want to place 3.26 mol NaOH(s) at the beginning of this DA. All we need now is a conversion factor relating moles of NaOH(s) to moles of H 3 PO 4 (aq). For this, we can use the balanced equation above. Remember, the coefficients in equations stand for moles, so: 3 mol NaOH(s) = 1 mol H 3 PO 4 (aq) If we’ve lined up the conversion factor correctly, everything should cancel out to give “mol Na 3 PO 4 (aq)” at the end. Everything checks out. Simply multiply the top and divide by the bottom and we’ll have our answer. Don’t forget to label your answer- this is very important in chemistry. = 1.09 mol H 3 PO 4 (aq)

7. 2H 2 O 2 (aq)  2H 2 O (g) + O 2 (g) ___________________________________ Example 2: mol  mol 1 mol O 2 (g) 2 mol H 2 O 2 (aq) _ _ _ _ 73.2 mol H 2 O 2 (aq) If 73.2 mol of hydrogen peroxide, H 2 O 2 (aq) decomposes according to the reaction above, how many moles of oxygen gas, O 2 (g), will be produced? Once again, this problem will involve the use of a DA. Remember, many (most) chemistry problems can be solved using dimensional analysis. We were given 73.2 mol H 2 O 2, so we’ll place that at the beginning of the DA. According to the equation, 2 mol of hydrogen peroxide will decompose to produce 1 mol of oxygen gas. So, 2 mol H 2 O 2 (aq)= 1 mol O 2 (g) As always, we need to check our work and, when all the units have cancelled, we are left with moles of O 2 (g). This is what we expected. To solve, we’ll multiply the top and divide by the bottom. = 36.6 mol O 2 (g)

8. Everything checks out. Multiply the top and divide by the bottom to find your answer. From the equation, we know that 12 mol CO 2 is produced by burning (combusting ) 2 mol of C 6 H 14, so: 12 mol CO 2 = 2 mol C 6 H 14 = 18 mol C 6 H 14 To begin, we’ll draw the framework for our DA. As usual, this is just a conversion between moles. You’ve been given 108 moles of carbon dioxide, CO 2, to start out. Place this at the head of the DA. As usual, we’ll double check our work by cancelling the units out. Remember: this helps you make sure that your conversion factor is lined up correctly. 2C 6 H 14 + 19O 2  12CO 2 + 14H 2 O ___________________________________ Example 3: mol  mol 2 mol C 6 H 14 12 mol CO 2 _ _ _ _ 108 mol CO 2 If this reaction produces 108 mol of CO 2, how many moles of hexane, C 6 H 14, were consumed to do so?

9. Although this conversion is slightly different from before, we will still need to use DA. This will be a 3-step DA, rather than just 2. 2C 6 H 14 + 19O 2  12CO 2 + 14H 2 O ___________________________________ Example 4: mol  mol  g 2 mol C 6 H 14 19 mol O 2 _ _ _ _ 1.33 mol O 2 If this reaction consumes 1.33 mol O 2, what mass of hexane, C 6 H 14, was consumed to do so? As usual, place the 1.33 mol O 2 at the head of the DA. We will work through the conversions until we reach grams of hexane. To get to hexane, we’ll need to use our equation. Remember, the equation provides relationships between different substances in terms of moles. From the equation, we see that 19 mol O 2 is consumed along with 2 mol C 6 H 14. We can conclude that: 19 mol O 2 = 2 mol C 6 H 14 This step in the DA converts mol O 2 to mol C 6 H 14, but we are looking for mass. Using the molar mass, we can add another step in the DA so we end in grams rather than moles C 6 H 14. We know that the molar mass of C 6 H 14 is 86.2 g/mol. This is the mass of 1 mole, so: 1 mol C 6 H 14 = 86.2 g C 6 H 14 Overall, we’ve changed one type of mole to another type and then gone back to grams. If we’ve set this up right, the dimensions (units) should cancel to give “g C 6 H 14.” Everything checks out. Multiply the top and divide by the bottom to find your answer. (Don’t forget your label!) _ _ _ _ 1 mol C 6 H 14 86.2 g C 6 H 14 = 12.1 g C 6 H 14

10. The only way we can convert between hexane and water is in moles. We’ll need to get to moles first if we want to use the equation. Use the molar mass: 1 mol C 6 H 14 = 86.2 g C 6 H 14 Theoretical yield problems are simple to understand. Essentially, you are being asked how much of a product (water) would theoretically be produced by burning a certain amount of reactant. 2C 6 H 14 + 19O 2  12CO 2 + 14H 2 O ___________________________________ Example 5: g  mol  mol 1 mol C 6 H 14 86.2 g C 6 H 14 _ _ _ _ 20.7 g C 6 H 14 If you are given 20.7g of hexane, C 6 H 14, and burn it in air, what is the theoretical yield of water in moles? _ _ _ _ 2 mol C 6 H 14 14 mol H 2 O Knowing this, we will simply set up a DA and solve for moles of water at the end. Place what you were given (20.7g C 6 H 14 ) at the beginning. We do not have any other information yet. Now, we have moles of hexane, but we want moles of water. We can use the equation to relate these two: 2 mol C 6 H 14 = 14 mol H 2 O If we’ve lined everything up right, all other units should cancel to give us the dimensions of our answer, mol H 2 O. Everything seems to be correct. Multiply the top and divide by the bottom to find the theoretical yield of water. = 1.68 mol H 2 O

11. This problem (above) is much the same as others we’ve seen. We’ll have to start with moles and convert them into grams of another substance. ___________________________________ Example 6: mol  mol  g 2 mol H 2 O 2 (aq) 1 mol O 2 (g) _ _ _ _ 383 mol O 2 (g) If you wanted to produce 383 mol of oxygen gas, how many grams of hydrogen peroxide would have to be decomposed? _ _ _ _ 1 mol H 2 O 2 (aq) 34.02 g H 2 O 2 (aq) = 26100 or 2.61e4 g H 2 O 2 (aq) 2H 2 O 2 (aq)  2H 2 O (g) + O 2 (g) We’ll have to start off with the 383 mol O 2 (g) given to us by the problem. Now, we know that there is only one way to convert between different substances: a balanced equation. So, from the equation above: 2 mol H 2 O 2 (aq) = 1 mol O 2 (g) We’re now in “mol H 2 O 2 ” so our final step should be to use the molar mass and convert to grams. 1 mol H 2 O 2 = 34.02 g H 2 O 2 Note: the (aq) and (g) symbols are indicators of the state of a substance, i.e. solid, liquid, gas, etc. They do not affect molar mass or conversions. Check your work by cancelling. The problem is asking for grams of hydrogen peroxide, so we should end in “g H 2 O 2 (aq).” Our conversions are lined up correctly. Solve the DA by multiplying the top and dividing by the bottom.

12. Our first goal should be to convert to moles. We have to use the equation (and moles) if we want to get to NaOH. 1 mol H 3 PO 4 = 98.00 g H 3 PO 4 Now that we’re in moles, we can use the equation above, so: 3 mol NaOH = 1 mol H 3 PO 4 Lastly, we need to turn moles NaOH back into grams. We used molar mass to get to moles, we can do so to return: 1 mol NaOH = 40.00 g NaOH Now, we need to check our work. This step becomes especially important for long DAs. After all, the more conversions you use, the more likely you might make a mistake. We’ve done well. Multiply the top and divide the bottom to find our final answer. = 60.0 g NaOH Unlike most of the previous problems, converting grams to grams (g  g) is the most common and useful skill taught in this unit. Remember, lab scales don’t read in moles. ___________________________________ Example 7: g  mol  mol  g 1 mol H 3 PO 4 98.00 g H 3 PO 4 _ _ _ _ 49.0 g H 3 PO 4 How many grams of sodium hydroxide would be required to neutralize (react with) 49.0g H 3 PO 4 ? _ _ _ _ 1 mol H 3 PO 4 3 mol NaOH 3NaOH(s) + H 3 PO 4 (aq)  3H 2 O(l) + Na3PO 4 (aq) 1 mol NaOH 40.00 g NaOH _ _ _ _ Despite the difference in conversions, the general pattern is still the same- we’ll start with the 49.0 g H 3 PO 4 we were given.

13. As before, we’re going to convert grams of one substance to grams of another. Although the equation is different, the method will be virtually the same. ___________________________________ Example 8: g  mol  mol  g 1 mol O 2 32 g O 2 _ _ _ _ 100. g O 2 How many grams of hexane, C 6 H 14, will burn in 100. grams of oxygen gas? _ _ _ _ 19 mol O 2 2 mol C 6 H 14 1 mol C 6 H 14 86.20 g C 6 H 14 _ _ _ _ 2C 6 H 14 + 19O 2  12CO 2 + 14H 2 O First, place the 100. g O 2 at the head of the DA. Use the molar mass to convert to moles. Remember: moles allow you to use the chemical equation. 1 mol O 2 = 32 g O 2 Next, use the equation to convert from moles of O 2 to moles of C 6 H 14. 19 mol O 2 = 2 mol C 6 H 14 Now, convert from moles of C 6 H 14 back to grams using the molar mass: 1 mol C 6 H 14 = 86.20 g C 6 H 14 Check your work and make sure you are left with grams of hexane, “g C 6 H 14 ” Solve by multiplying the top and dividing by the bottom. = 28.4 g C 6 H 14

14. ________________________________ Practice and Review Problems (I): 1 mol C 6 H 14 86.20 g C 6 H 14 _ _ _ _ 50.0 g C 6 H 14 Given 50.0 of hexane, C 6 H 14, determine the following: a.) O 2 consumedb.) CO 2 producedc. H 2 O produced _ _ _ _ 2 mol C 6 H 14 19 mol O 2 1 mol O 2 32.0 g O 2 _ _ _ _ 2C 6 H 14 + 19O 2  12CO 2 + 14H 2 O ________________________________ 1 mol C 6 H 14 86.20 g C 6 H 14 _ _ _ _ 50.0 g C 6 H 14 _ _ _ _ 2 mol C 6 H 14 12 mol CO 2 1 mol CO 2 44.01 g CO 2 _ _ _ _ ________________________________ 1 mol C 6 H 14 86.20 g C 6 H 14 _ _ _ _ 50.0 g C 6 H 14 _ _ _ _ 2 mol C 6 H 14 14 mol H 2 O 1 mol H 2 O 18.02 g H 2 O _ _ _ _ a.) b.) c.) 176 g O 2 153 g CO 2 73 g H 2 O

15. Coincidence? Perhaps not! 2C 6 H 14 + 19O 2  12CO 2 + 14H 2 O 176 g 153 g 73 g 50 g If you take all of the measurements we calculated in the prior problem and place them below the equation, you will notice something very interesting. (Remember, we used 50 g of hexane, C 6 H 14, as the starting point for each of these as well.) If you add up all the reactant and do the same for the products………… They both total up to a mass of 226 g! Remember, THIS is the Law of Conservation of Mass/Matter at work. “What goes in must come out!” + + 226 g “In” 226 g “Out”=

16. ________________________________ Practice and Review Problems (II): 1 mol Al 26.98 g Al _ _ _ _ 90.2 g Al Given 90.2 g of aluminum powder, Al.: a.) O 2 consumedb.) Al 2 O 3 produced _ _ _ _ 4 mol Al 3 mol O 2 1 mol O 2 32.0 g O 2 _ _ _ _ 4Al + 3O 2  2Al 2 O 3 a.) b.) 80.2 g O 2 170. g Al 2 O 3 ________________________________ 1 mol Al 26.98 g Al _ _ _ _ 90.2 g Al _ _ _ _ 4 mol Al 2 mol Al 2 O 3 1 mol Al 2 O 3 101.96 g Al 2 O 3 _ _ _ _