1. Ppt 11 Plan (PS5, 1-11 material) 1.Meaning of Coefficients in a Balanced Equation  Ratio of reaction (in FU or moles), not actual amounts  “Standard” vs. nonstandard (difference between a product and leftover reactant) 2.Nanoscopic Interpretation (FUs) [Not stressed in Tro?]  Predict the equation from picture  Predict final picture from initial (pic) and equation 3.Macroscopic Interpretation (moles [of FUs] ) (Tro, 4.2)  Mole ratios as 4 th “interconversion factor”  mol A  mol B; mol A, B  g B, A; g A  g B  Problems on old Stoichiometry Quiz 4.How to Balance an Equation (Tro, 3.10) 1Ppt 11

2. Example—Nanoscopic Interpretation of a balanced equation “For every N 2 molecule that reacts three molecules of H 2 react with it to form two NH 3 molecules.” – Ratio is 3 H 2 lost : 1 N 2 lost : 2 NH 3 formed N 2 + 3 H 2  2 NH 3 means what? Does this tell you how much N 2 you start with? Does this tell you how much H 2 you start with? Does this tell you how much NH 3 is made? NO!! 2Ppt 11 Note: The same number of atoms are represented on the right side of the arrow as the left.

3. Coefficients Represent a ratio only (not actual amounts) Ratio is only of FU or moles (not grams!) Only when chemical change (rxn) takes place – Not the ratio of amounts present at the beginning – Not the ratio of amounts present at end If I could, I’d define something called an “equation unit” of reaction: – Smallest amount of reaction that could possibly occur. – Coeffs represent the exact number of each FU used and made when one “equation unit” of reaction occurs 3Ppt 11

4. “Standard” Balanced Equation Coefficients are in the lowest whole number ratio The same substance appears only once on each side of the equation – If it’s on both sides, it didn’t actually change! – Leftover reactant is not a product! 4Ppt 11 These equations are balanced! They’re just not in standard form. Technically balanced, but gives wrong ratio and implies H 2 is made!

5. Application of Ideas I—Nanoscopic Pics/Interpretation See PS4 & 5 Practice Worksheet, Problems #7 and #8 5Ppt 11 7. Reaction of A (open spheres) with B (black spheres) is shown schematically in the following diagram: Which equation best describes the stoichiometry of the reaction (the ratio in which substances react and form when the reaction takes place)? (a) A 2 + 2 B → A 2 B 2 (b) 10 A + 5 B 2 → 5 A 2 B 2 (c) 2 A + B 2 → A 2 B 2 (d) 5 A + 5 B 2 → 5 A 2 B 2 Ans. (c) It gives the ratio (lowest whole number). Equation is not meant to represent “actual amounts”.

6. Application of Ideas I—Nanoscopic Pics/Interpretation See PS4 & 5 Practice Worksheet, Problems #7 and #8 6Ppt 11 8. If the diagrams below represent a reaction that occurred in a closed container according to the equation: 2 NO(g) + O 2 (g) → 2 NO 2 (g), what would be left in the box after the reaction has gone as completely as possible? DONE ON OVERHEAD

7. And then: 2 NO + O 2  2 NO 2 (ratio reduced) The ratio in which the reactants “reacted” is 2:1, not 6:5! O 2 is not a product! Application of Ideas I—Nanoscopic Pics/Interpretation #8 Follow up. NOTE: 6 NO + 5 O 2  6 NO 2 + 2 O 2 is not an equation in standard form! “Simplify” to: 6 NO + 3 O 2  6 NO 2 (O 2 ’s “cancelled”) 7Ppt 11

8. PS Sign-Posting The concepts and skills related to problems 1-4 on PS5 were covered in the prior section of this PowerPoint. Give those problems a try now! 8Ppt 11

9. Application of Ideas II—Macroscopic Interpretation “For every 1 mole of N 2 (molecules) that react, 3 moles of H 2 (molecules) react with them to form 2 moles of NH 3 (molecules) – Ratio is 3 moles of H 2 lost : 1 mole of N 2 lost : 2 moles of NH 3 formed N 2 + 3 H 2  2 NH 3 also means: 9Ppt 11

10. Thus….Mole RATIOS Can Be Made (and Used!) From: N 2 + 3 H 2  2 NH 3 you can create….mole ratio “conversion factors” 10Ppt 11 And reciprocals:

11. Bertrand Applet—Applying “Mole ratio” Idea to Chemical Reactions Ppt 1111 Try out the following applet to see if you really understand the meaning of a balanced chemical equation (and the difference between an equation and a chemical reaction)! http://web.mst.edu/~gbert/reactor/Areactor.html

12. Take home lessons from web exercise (prior slide) The amount of a reactant that is present to begin with is not necessarily equal to the amount that reacts – Some might be left over (not reacted)! The amount of a product that is present at the end is not necessarily equal to the amount that formed – Some might have been present to start with! Coefficient ratios apply only to the “change” in R’s or P’s Use “I C F” (Initial, Change, Final) table to help see this 12Ppt 11

13. See Stoichiometry Quiz (Chemistry 121 Quiz Used as PRACTICE WORKSHEET on "Early" Stoichiometry) 1. (8 pts) Given the following chemical equation: P 4 O 10 + 5 CCl 4  5 CO 2 + 4 PCl 3 + 4 Cl 2 a.How many moles of CO 2 will be formed if 3 moles of P 4 O 10 react? b.How many moles of P 4 O 10 would be used up if 1.9 moles of Cl 2 were produced? c.How many moles of PCl 3 form if 2.4 moles of CCl 4 react? d.How many moles of CO 2 will be formed if 0.247 moles of P 4 O 10 react? e.How many grams of Cl 2 would be formed if 3.2 moles of P 4 O 10 were reacted? f.If 32.6 g of CCl 4 reacts, how many grams of Cl 2 form? 13Ppt 11

14. Calculating Mass of Reactants and Products Reacted or formed! 14Ppt 11

15. PS Sign-Posting The concepts and skills related to problems 5-7 on PS5 were covered in the prior section of this PowerPoint *. Give those problems a try now! * Problems 12a and 12b have also been covered, but the term “limiting reactant”, which is in part c of this problem, will be covered in Ppt12. 15Ppt 11

16. How To Balance a Chemical Equation-I Balancing Means “Adding Coefficients” – Not subscripts!! Must ALREADY KNOW the substances that are reactants and products – The FORMULAS must be determined FIRST! – Write the formulas of the reactants on the left of the arrow and those of the products on the right Method of “Committed Coefficients” Idea – (next slide) 16Ppt 11

17. How To Balance a Chemical Equation-II “Method of Committed Coefficients” 1.Pick one of the most complex-looking formulas and make the coefficient a “1”. – That coefficient is now “committed” 2.Find an atom type that occurs only in that formula (since it is “committed”), and in only ONE formula on the other side of the arrow (if possible). – “Balance” that atom type by adding a coefficient (to make total atoms on each side equal). – Now you have TWO committed coefficients! 3.Repeat the steps above, always looking first for atom types that appear in the fewest number of formulas  Always leave the formulas of ELEMENTS for last! 17Ppt 11

18. Method of Committed Coefficients (cont.) 4.If you get to a point where you need to use a FRACTION to get the number of atoms you “need”, then use the fraction as a coefficient! – At the end, multiply through all coefficients by the denominator of your fraction to end up with a whole number ratio. – If you prefer to avoid fractions, you can “start over” with a “2” (or “3”) in place of the original “1” instead of doing “Step 4” here. 5.Important tip: – When counting up atoms before placing your next coefficient, only count those atoms that come from formulas with committed coefficients. i.e., Don’t just count up all the atoms of X on both sides of the equation to start off with! This will assume that all the coefficients are “1” when most of the time that will not be the case once the equation is balanced. 18Ppt 11

19. Example(s) Al + NH 4 ClO 4 → Al 2 O 3 + AlCl 3 + NO + H 2 O Al + 1 NH 4 ClO 4 → Al 2 O 3 + AlCl 3 + NO + H 2 O 1 Look at right side: N, H, and Cl are “isolated”; O is in many places O’s: on LEFT, 4  Need ONE more 1 3 Al + 3 NH 4 ClO 4 → 1 Al 2 O 3 + 1 AlCl 3 + 3 NO + 6 H 2 O 12 on RIGHT, 3 committed (don’t count Al 2 O 3 !)  coefficient of 1/3 for Al 2 O 3 Al’s: on RIGHT, 2/3 + 1/3 = 1  on LEFT, commit a “1” All coeffs done. Multiply whole equation (all coeffs) by 3: 19Ppt 11

20. Balancing Equations Examples/Practice (handout) 1.___ Cu 2 O + ___ Cu 2 S → ___ Cu + ___ SO 2 2.___ HCl + ___ Al(OH) 3 → ___ AlCl 3 + ___ H 2 O 3.___ CH 3 OH + ___ O 2 → ___ HCHO + ___ H 2 O 4.___ P + ___ Fe 2 O 3 → ___ P 4 O 10 + ___ Fe 5.___ Pb + ___ PbO 2 + ___ H 2 SO 4 → ___ PbSO 4 + ___ H 2 O 6.___ PbO + ___ PbS → ___ Pb + ___ SO 2 20Ppt 11

21. PS Sign-Posting The concepts and skills related to problems 8-11 on PS5 were covered in the prior section of this PowerPoint. Give those problems a try now! 21Ppt 11