1 / 16

SKILLS Project

SKILLS Project. Reaction Stoichiometry and Theoretical Yield. 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:

rosina
Télécharger la présentation

SKILLS Project

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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 58.44g NaCl mol  g 1 mol NaCl 50g NaCl 1 mol NaCl g  mol 58.44g NaCl The sameconversion 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. Example 1: mol  mol 3NaOH(s) + H3PO4(aq)  3H2O(l) + Na3PO4(aq) Given 3.26 mol NaOH(s), calculate the number of moles of H3PO4(aq) that will be consumed. ___________________________________ ____ 1 mol H3PO4(aq) 3.26 mol NaOH(s) 3 mol NaOH(s) • = 1.09 mol H3PO4(aq) • As you learned in the “Basic DA” unit, we’ll want to place 3.26 mol NaOH(s) at the beginning of this DA. • If we’ve lined up the conversion factor correctly, everything should cancel out to give “mol Na3PO4(aq)” at the end. • Remember, the coefficients in equations stand for moles, so: • 3 mol NaOH(s) = 1 mol H3PO4(aq) • All we need now is a conversion factor relating moles of NaOH(s) to moles of H3PO4(aq). For this, we can use the balanced equation above. • 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. • 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.

  7. Example 2: mol  mol 2H2O2(aq)  2H2O(g) + O2(g) If 73.2 mol of hydrogen peroxide, H2O2(aq) decomposes according to the reaction above, how many moles of oxygen gas, O2(g), will be produced? ___________________________________ ____ 1 mol O2(g) 73.2 mol H2O2(aq) 2 mol H2O2(aq) = 36.6 mol O2(g) • As always, we need to check our work and, when all the units have cancelled, we are left with moles of O2(g). This is what we expected. • To solve, we’ll multiply the top and divide by the bottom. • Once again, this problem will involve the use of a DA. Remember, many (most) chemistry problems can be solved using dimensional analysis. • According to the equation, 2 mol of hydrogen peroxide will decompose to produce 1 mol of oxygen gas. So, • 2 mol H2O2(aq)= 1 mol O2(g) • We were given 73.2 mol H2O2, so we’ll place that at the beginning of the DA.

  8. Example 3: mol  mol 2C6H14 + 19O2 12CO2 + 14H2O If this reaction produces 108 mol of CO2, how many moles of hexane, C6H14, were consumed to do so? ___________________________________ ____ 2 mol C6H14 108 mol CO2 12 mol CO2 = 18 mol C6H14 • As usual, we’ll double check our work by cancelling the units out. Remember: this helps you make sure that your conversionfactor is lined up correctly. • To begin, we’ll draw the framework for our DA. As usual, this is just a conversion between moles. • From the equation, we know that 12 mol CO2 is produced by burning (combusting ) 2 mol of C6H14, so: • 12 mol CO2 = 2 mol C6H14 • Everything checks out. Multiply the top and divide by the bottom to find your answer. • You’ve been given 108 moles of carbon dioxide, CO2, to start out. Place this at the head of the DA.

  9. Example 4: mol  mol  g 2C6H14 + 19O2 12CO2 + 14H2O If this reaction consumes 1.33 mol O2, what mass of hexane, C6H14, was consumed to do so? ___________________________________ ____ ____ 86.2 g C6H14 2 mol C6H14 1.33 mol O2 19 mol O2 1 mol C6H14 = 12.1 g C6H14 • Everything checks out. Multiply the top and divide by the bottom to find your answer. (Don’t forget your label!) • To get to hexane, we’ll need to use our equation. Remember, the equation provides relationships between different substances in terms of moles. • 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 C6H14.” • As usual, place the 1.33 mol O2 at the head of the DA. We will work through the conversions until we reach grams of hexane. • We know that the molar mass of C6H14 is 86.2 g/mol. This is the mass of 1 mole, so: • 1 mol C6H14 = 86.2 g C6H14 • This step in the DA converts mol O2 to mol C6H14, 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 C6H14. • 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. • From the equation, we see that 19 mol O2 is consumed along with 2 mol C6H14. We can conclude that: • 19 mol O2 = 2 mol C6H14

  10. Example 5: g  mol  mol 2C6H14 + 19O2 12CO2 + 14H2O If you are given 20.7g of hexane, C6H14, and burn it in air, what is the theoretical yield of water in moles? ___________________________________ ____ ____ 14 mol H2O 1 mol C6H14 20.7 g C6H14 86.2 g C6H14 2 mol C6H14 = 1.68 mol H2O • If we’ve lined everything up right, all other units should cancel to give us the dimensions of our answer, mol H2O. • Place what you were given (20.7g C6H14) at the beginning. We do not have any other information yet. • Knowing this, we will simply set up a DA and solve for moles of water at the end. • Now, we have moles of hexane, but we want moles of water. We can use the equation to relate these two: • 2 mol C6H14 = 14 mol H2O • 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. • Everything seems to be correct. Multiply the top and divide by the bottom to find the theoretical yield of water. • 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 C6H14 = 86.2 g C6H14

  11. Example 6: mol  mol  g 2H2O2(aq)  2H2O(g) + O2(g) If you wanted to produce 383 mol of oxygen gas, how many grams of hydrogen peroxide would have to be decomposed? ___________________________________ ____ ____ 34.02 g H2O2(aq) 2 mol H2O2(aq) 383 mol O2(g) 1 mol O2(g) 1 mol H2O2(aq) = 26100 or 2.61e4 g H2O2(aq) • We’re now in “mol H2O2” so our final step should be to use the molar mass and convert to grams. • 1 mol H2O2 = 34.02 g H2O2 • 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. • Our conversions are lined up correctly. Solve the DA by multiplying the top and dividing by the bottom. • We’ll have to start off with the 383 mol O2(g) given to us by the problem. • Check your work by cancelling. The problem is asking for grams of hydrogen peroxide, so we should end in “g H2O2(aq).” • Now, we know that there is only one way to convert between different substances: a balanced equation. So, from the equation above: • 2 mol H2O2(aq) = 1 mol O2(g) • 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.

  12. Example 7: g  mol  mol  g 3NaOH(s) + H3PO4(aq)  3H2O(l) + Na3PO4(aq) How many grams of sodium hydroxide would be required to neutralize (react with) 49.0g H3PO4? ___________________________________ ____ ____ ____ 3 mol NaOH 40.00 g NaOH 49.0 g H3PO4 1 mol H3PO4 1 mol H3PO4 98.00 g H3PO4 1 mol NaOH = 60.0 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. • 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 H3PO4 = 98.00 g H3PO4 • 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 • Despite the difference in conversions, the general pattern is still the same- we’ll start with the 49.0 g H3PO4 we were given. • Now that we’re in moles, we can use the equation above, so: • 3 mol NaOH = 1 mol H3PO4 • 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. • We’ve done well. Multiply the top and divide the bottom to find our final answer.

  13. Example 8: g  mol  mol  g 2C6H14 + 19O2 12CO2 + 14H2O How many grams of hexane, C6H14, will burn in 100. grams of oxygen gas? ___________________________________ ____ ____ ____ 2 mol C6H14 86.20 g C6H14 100. g O2 1 mol O2 19 mol O2 32 g O2 1 mol C6H14 = 28.4 g C6H14 • Now, convert from moles of C6H14 back to grams using the molar mass: • 1 mol C6H14 = 86.20 g C6H14 • 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. • Next, use the equation to convert from moles of O2 to moles of C6H14. • 19 mol O2 = 2 mol C6H14 • Solve by multiplying the top and dividing by the bottom. • Use the molar mass to convert to moles. Remember: moles allow you to use the chemical equation. • 1 mol O2 = 32 g O2 • Check your work and make sure you are left with grams of hexane, “g C6H14” • First, place the 100. g O2 at the head of the DA.

  14. Practice and Review Problems (I): 2C6H14 + 19O2 12CO2 + 14H2O Given 50.0 of hexane, C6H14, determine the following: a.) O2 consumed b.) CO2 produced c. H2O produced 73 g H2O 153 g CO2 176 g O2 ________________________________ ____ ____ ____ 19 mol O2 32.0 g O2 50.0 g C6H14 1 mol C6H14 a.) 2 mol C6H14 86.20 g C6H14 1 mol O2 ________________________________ ____ ____ ____ 12 mol CO2 44.01 g CO2 50.0 g C6H14 1 mol C6H14 b.) 2 mol C6H14 86.20 g C6H14 1 mol CO2 ________________________________ ____ ____ ____ 14 mol H2O 18.02 g H2O 50.0 g C6H14 1 mol C6H14 c.) 2 mol C6H14 86.20 g C6H14 1 mol H2O

  15. Coincidence? Perhaps not! 2C6H14 + 19O2 12CO2 + 14H2O + + 73 g 50 g 153 g 176 g = 226 g “Out” 226 g “In” • If you add up all the reactant and do the same for the products………… • (Remember, we used 50 g of hexane, C6H14, as the starting point for each of these as well.) • 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. • 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!”

  16. Practice and Review Problems (II): 4Al + 3O2 2Al2O3 Given 90.2 g of aluminum powder, Al.: a.) O2 consumed b.) Al2O3 produced 170. g Al2O3 80.2 g O2 ________________________________ ____ ____ ____ 3 mol O2 32.0 g O2 90.2 g Al 1 mol Al a.) 4 mol Al 26.98 g Al 1 mol O2 ________________________________ ____ ____ b.) ____ 2 mol Al2O3 101.96 g Al2O3 90.2 g Al 1 mol Al 4 mol Al 26.98 g Al 1 mol Al2O3

More Related