1. Ppt 11 Plan (PS5, 1-11 material)
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)
Nanoscopic Interpretation (FUs) [Not stressed in Tro?]
Predict the equation from picture
Predict final picture from initial (pic) and equation
Macroscopic Interpretation (moles [of FUs]) (Tro, 4.2)
Mole ratios as 4th “interconversion factor”
mol A  mol B; mol A, B  g B, A; g A  g B
Problems on old Stoichiometry Quiz
How to Balance an Equation (Tro, 3.10)
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2. Example—Nanoscopic Interpretation of a balanced equation
Note: The same number of atoms are represented on the right side of the arrow as the left.
N H2  2 NH3 means what?
“For every N2 molecule that reacts three molecules of H2 react with it to form two NH3 molecules.”
Ratio is 3 H2 lost : 1 N2 lost : 2 NH3 formed
NO!!
Does this tell you how much N2 you start with?
Does this tell you how much H2 you start with?
NO!!
Does this tell you how much NH3 is made?
NO!!
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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
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4. “Standard” Balanced Equation
Coefficients are in the lowest whole number ratio
These equations are balanced! They’re just not in standard form.
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!
Technically balanced, but gives wrong ratio and implies H2 is made!
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5. Application of Ideas I—Nanoscopic Pics/Interpretation
See PS4 & 5 Practice Worksheet, Problems #7 and #8
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) A2 + 2 B → A2B2
(b) 10 A + 5 B2 → 5 A2B2
(c) 2 A + B2 → A2B2
(d) 5 A + 5 B2 → 5 A2B2
Ans. (c) It gives the ratio (lowest whole number). Equation is not meant to represent “actual amounts”.
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6. Application of Ideas I—Nanoscopic Pics/Interpretation
See PS4 & 5 Practice Worksheet, Problems #7 and #8
8. If the diagrams below represent a reaction that occurred in a closed container according to the equation: 2 NO(g) + O2(g) → 2 NO2(g), what would be left in the box after the reaction has gone as completely as possible?
DONE ON OVERHEAD
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7. Application of Ideas I—Nanoscopic Pics/Interpretation
#8 Follow up.
NOTE:
6 NO O2  6 NO O2
is not an equation in standard form!
“Simplify” to: 6 NO O2  6 NO2 (O2’s “cancelled”)
And then: NO O2  2 NO2 (ratio reduced)
The ratio in which the reactants “reacted” is 2:1, not 6:5!
O2 is not a product!
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8. PS Sign-Posting
The concepts and skills related to problems 1-3 on Written PS5 (and some Mastering Qs) were covered in the prior section of this PowerPoint. Give those problems a try now!
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9. Application of Ideas II—Macroscopic Interpretation
N H2  2 NH3 also means:
“For every 1 mole of N2 (molecules) that react, 3 moles of H2 (molecules) react with them to form 2 moles of NH3 (molecules)
Ratio is 3 moles of H2 lost : 1 mole of N2 lost : 2 moles of NH3 formed
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10. Thus….Mole RATIOS Can Be Made (and Used!)
From: N H2  2 NH3
you can create….mole ratio “conversion factors”
And reciprocals:
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11. See Stoichiometry Quiz (Chemistry 121 Quiz Used as PRACTICE WORKSHEET on "Early" Stoichiometry)
1. (8 pts) Given the following chemical equation:
P4O CCl4  CO PCl Cl2
How many moles of CO2 will be formed if 3 moles of P4O10 react?
How many moles of P4O10 would be used up if 1.9 moles of Cl2 were produced?
How many moles of PCl3 form if 2.4 moles of CCl4 react?
How many moles of CO2 will be formed if moles of P4O10 react?
How many grams of Cl2 would be formed if 3.2 moles of P4O10 were reacted?
If 32.6 g of CCl4 reacts, how many grams of Cl2 form?
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12. Solution to (f) on prior worksheet
If 32.6 g of CCl4 reacts according to the equation below, how many grams of Cl2 will form?
P4O CCl4  CO PCl Cl2
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13. Calculating Mass of Reactants and Products
Reacted or formed!
Reacted or formed!
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14. Bertrand Applet—Applying “Mole ratio” Idea to Chemical Reactions
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)!
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15. 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
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16. PS Sign-Posting
The concepts and skills related to problem 4 and 7* on Written PS5 (and more problems in Mastering) were covered in the prior section of this PowerPoint. Give those problems a try now!
* Problem 7a and 7b have been covered, but the term “limiting reactant”, which is in part c of Q7, will be covered in Ppt12.
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17. 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)
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18. How To Balance a Chemical Equation-II “Method of Committed Coefficients”
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!
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19. Method of Committed Coefficients (cont.)
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.
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.
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20. Example(s) Al + NH4ClO4 → Al2O3 + AlCl3 + NO + H2O 1 1
Look at right side: N, H, and Cl are “isolated”; O is in many places 1
Al + 1 NH4ClO4 → Al2O AlCl NO + H2O 1 2
O’s: on LEFT, 4
on RIGHT, 3 committed (don’t count Al2O3!)
 Need ONE more
 coefficient of 1/3 for Al2O3
Al’s: on RIGHT, 2/3 + 1/3 = 1
 on LEFT, commit a “1”
All coeffs done. Multiply whole equation (all coeffs) by 3:
3 Al + 3 NH4ClO4 → 1 Al2O3 + 1 AlCl3 + 3 NO + 6 H2O
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21. Balancing Equations Examples/Practice (handout)
___ Cu2O + ___ Cu2S → ___ Cu + ___ SO2
___ HCl + ___ Al(OH)3 → ___ AlCl3 + ___ H2O
___ CH3OH + ___ O2 → ___ HCHO + ___ H2O
___ P + ___ Fe2O3 → ___ P4O10 + ___ Fe
___ Pb + ___ PbO2 + ___ H2SO4 → ___ PbSO4 + ___ H2O
___ PbO + ___ PbS → ___ Pb + ___ SO2
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22. PS Sign-Posting
The concepts and skills related to problems 5 and 6 on Written PS5 (and more problems in Mastering) were covered in the prior section of this PowerPoint. Give those problems a try now!
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