September 18, reaction stoichiometry.notebook. May 18 10:07 AM. Sep 13 8:55 AM REACTION STOICHIOMETRY

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1 REACTION STOICHIOMETRY COMPOSITION STOICHIOMETRY: The mass relationships of elements in a compound REACTION STOICHIOMETRY : the mass relationships between products and reactants May 18 10:07 AM Sep 13 8:55 AM 1

2 Molar Ratios Consider the following balanced equation: 2H 2 O + Ca > Ca(OH) 2 + H 2 The coefficients tell us the ratios of the amount of compounds in the reaction Where there is no coefficient, we understand 1 May 1 1:54 PM All types of reaction stoichiometry calculations require knowing the molar ratios. Molar ratio: a conversion factor that relates the amounts, in moles, of any two substances involved in a chemical reaction. May 1 2:01 PM 2

3 2H 2 O + Ca > Ca(OH) 2 + H 2 For every 2 molecules or parts or moles of water we get 1 molecule, or part or mole of cacium hydroxide. comparing them in a ratio: 2:1 When we discuss the amount in moles, this ratio is known as the molar ratio. May 1 1:58 PM Molar ratio, cont'd We get the molar ratio from the balanced equation. the decomposition of aluminum oxide yields aluminum metal and oxygen gas. What are all of the possible molar ratios? May 1 2:04 PM 3

4 Al 2 O 3 : Al 2Al 2 O 3 > 4Al (s) + 3O 2 (g) Al 2 O 3 : O 2 Al :O 2 May 1 2:06 PM Sep 13 10:39 AM 4

5 Type1. All of the problems we deal with can be summed up into 4 types: The given and unknown quantities are both in moles (mole to mole calculations) amnt of given substance amount unknown substance ( in moles) ( in moles) May 18 10:17 AM Process: write a balanced equation with correct formulas set up ratio of known to unknown based on balanced equation mole ratios complete the proportion answer = # moles May 24 7:58 AM 5

6 May 24 8:14 AM Sep 13 9:38 AM 6

7 az Sep 14 9:02 AM Sep 14 9:09 AM 7

8 Sep 14 9:30 AM Sep 17 8:53 AM 8

9 Sep 17 9:00 AM 7 Type 2 Process: The given is in moles and the unknown is a mass expressed in grams write a balanced equation with correct formulas set up ratio of known to unknown based on balanced equation mole ratios complete the proportion answer = # moles convert unknown from moles to grams May 17 11:29 AM 9

10 Example: page 306 Photosynthesis Plants use energy from the sun, Water and Carbon Dioxide to make glucose. What mass, in grams, of glucose is produced when 3.0 mol of water react with enough carbon dioxide? 1. Write a balanced equation with correctly written formulas 6H 2O + 6CO 2 C 6H 12O O 2 2. determine the given and unknown May 8 10:45 AM Example: page determine the molar ratio of known to unknown in the equation 6H 2 O + 6CO 2 C 6 H 12 O O 2 4. set up a proportion between molar ratio and the given amounts in the problem May 8 1:28 PM 10

11 Sep 17 9:28 AM May 25 7:59 AM 11

12 Type 3 The given is in grams (mass) The unknown is an amount in moles Process: write a balanced equation with correct formulas convert known to moles #moles = mass (g) molar mass set up ratio of known to unknown based on balanced equation complete the proportion answer = # moles May 7 11:05 AM Example: page 309 Manufacture of nitric acid is the catalytic oxidation of ammonia Step 1: Write balanced equation 4NH 3 (g) + 5O 2(g) 4NO (g) + 6 H 2O (g) Step 2: determine given and unknown Given mass NH 3 = 824g? # mole NO =? # mole H 2O=? Step 3: determine # moles in the given mass molar mass = # moles g g/mole = mole 824 g g/mole = ex pg

13 Example: page 309 4NH 3 (g) + 5O 2(g) 4NO (g) + 6 H 2O (g) 4molNH 3 = 4mol NO x 4x = 4(48.36) mol NO x = 48.4 mol NO 4molNH 3 = 6mol H 2 O mol NH 3 4x = 6(48.36) mole H 2 O x = 72.5 mol H 2 O x May 29 9:14 AM Na Cl Given Type 4 The given is expressed in grams and the unknown is asked for in grams. Cannot form ratios in stoichometry using grams. Sn(s) + 2HF mass HF= 30.00g SnF 2 (s)+ H 2 (g)? mass SnF 2 =? calculate the molar mass of compounds in question HF= 1.01g/mol g/mol = g/mol SnF 2 = g/mol g/mol= g/mol mass molar mass =#mol How many moles of HF do we have? 30.00g HF g/mol = 1.5 mol HF May 17 11:54 AM 13

14 TYPE 4 1. Balance equation 2. Form molar ratio between known and unknown convert given mass moles Calc. # moles of unknown convert mole of unknown grams May 27 9:50 AM what is the molar ratio of given compound to unknown compound in the balanced equation? 2mol HF 1 molsnf 2 Next, form a proportion using the actual amount of given and the molar ratio from the balanced equation 2mol HF = 1.5 mol HF 1 molsnf 2 x x = 0.75 mol SnF 2 May 21 12:01 PM 14

15 Finally, convert the # of moles to mass mass = # moles. molar mass 0.75 mol SnF 2 x g/mol = g SnF 2 with significant figures = 118 g SnF 2 May 26 7:23 AM Recap: Type I: mole to mole Type 2: Type#1 ; then convert result g to mol Type 3: convert known g mol + Type I Type 4: convert known g mol + Type I; convert result mol to g Sep 18 9:17 AM 15

16 Sep 18 9:19 AM 1.5 C sugar +2½ C. flour 100 cookies What if I only have 1 cup of flour? how many cookies should I get? 2 1/2 C. flour = 1 c flour 100 cookies? 100 cookies (1) = 2.5 (?)? = 40 cookies May 18 11:05 AM 16

17 Suppose we don't have enough of either sugar or flour to make a full recipe is there a way to predict how many cookies we will get? Will one ingredient be more important than the other? Consider.65 cup of sugar and 1 cup of flour which one determines how many cookies? May 19 12:12 PM lets go back to our original "recipe" 1.5 C sugar + 2½ C. flour 100 cookies form 2 mole ratiosone between the product and each of the reactants: 1.5 C sugar =.65 C sugar 2½ C. flour 1 C. flour 100cookies x 100 cookies x x = cookies What this shows us is that the flour runs out before the sugarthe flour limits how much product we get. (Our limiting reagent) May 19 1:04 PM 17

18 So, Theoretically, we should get 40 cookies We bake the cookies and get 30 cookies! What happened? Quantitative analysis We won't be calculating a % error, but a % yield. What percentage of 40 is 30? actual yield x 100% = percent yield experimental 30/40 x 100% = 75% We got 75% of the cookies we expected to get. May 19 12:11 PM so, how does this relate to chemistry? Type 1: mole to mole equations In this type of calculation, we are given a reaction, usually in words, a known amount of a compound or element and asked to find the amount of product produced, or which reactant limited the reaction. May 19 11:26 AM 18

19 Example: ( page 305 of your book) In a spacecraft, the carbon dioxide exhaled by the astronauts can be removed by its reaction with lithium hydroxide, LiOH, according to the following chemical equation: write a balanced equation: (which is nicely done for us) CO2 (g) + 2 LiOH(s) Li2CO3(s) + H2O (l) May 19 11:40 AM known amt of H20=3moles 6 6 CO 2 + 6H 2 O C 6 H 12 O ? Mass of glucose =? Mole ratio from balanced equation 6 mole of H2O 1 mole C6H12O6 From this, set up a ratio : 6 mole of H2O = 3mol H2O 1 mole C6H12O6 x 6x = 3 mol C6H12O6 x =.5 mol C6H12O6 May 27 10:46 AM 19

20 Next: convert the amount (in moles) to the mass of glucose ( in grams) To do this we need to know the molar mass of glucose C 6H 12O 6 C: 6 x = 72.06g/mole H: 12 x 1.01 = 12.01g/mole O: 6 x 16 = 96. g/mole 180 g /mole May 27 10:53 AM 6 CO 2 + 6H 2 O C 6 H 12 O Apr 4 1:29 PM 20

21 ? Known.489 mol P mass O2 =? 4 mol P =.489 mol P 5mol O2 x 4x =.489x(5mol O2) molar mass O2 = 32 g/ mol x =.59 mol O2 mass O2 =.59 x 32 g/mol May 27 12:31 PM known?.489mol P mass P402=? 4 mol P =.489 mol P 1 mol P 4O 10 x 4 x = 1 mol P4O10 x =.25 mole P4O10 =.25 mol (283 g/mol) = g May 28 2:11 PM 21

22 Na2CO3+2HNO3 2NaNO3 + CO2 + H2O given? mass Na2CO3 = 100g amt of NaNO3 =? molar mass NaNO3 Na2CO3 molar mass of 1xNa = g/mol 2 x Na = g/ mole 1x N = l4.01 g/mol 1 x C = g/ mole 3 x O = 48 g/mol 3 x16 = 48 g/mol mol 85 g/ 100 g 85 g/mol = mole 106 g/mol 2 mol NaNO3 = mole NaNO3 1 mol Na2CO3 x x = 1.176/2 mol Na2CO3 =.588 mol Na2CO3 May 29 10:47 AM Apr 18 12:50 PM 22

23 May 15 11:01 AM Apr 18 12:52 PM 23

24 Apr 24 1:10 PM Apr 24 1:16 PM 24

25 Apr 24 1:26 PM Theoretical Yield:Ideal Calculations give us an amount of product we should get. Actual yield Amount of product actually obtained in an experiment. May 16 10:34 AM 25

26 Percentage yield: Actual Yield x 100% theoretical yield May 17 12:40 PM Exa Ideal: g mple : Actual: 23.72g % yield: g x 100% g 85.88% May 17 10:45 AM 26

27 May 15 10:36 AM May 15 10:50 AM 27

28 May 15 10:45 AM May 15 10:53 AM 28

29 May 15 10:53 AM Chapter 9 REACTION STOICHIOMETRY May 27 9:26 AM 29

30 Practice problems from chapter 3 If one dozen eggs weight 16.0 ounces, how much does one egg weigh? Apr 30 1:08 PM Apr 3 12:54 PM 30

31 Apr 3 12:52 PM Apr 3 1:01 PM 31

32 Apr 3 1:05 PM Na Cl Na Cl Na Cl Na Cl Na Cl Na Cl Na Cl Na Cl Na Cl Apr 3 1:11 PM 32

33 Sep 21 10:39 AM Sep 21 10:48 AM 33