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Excess Reactant. Visualizing Limiting Reactant 1 23 4 5 6 7 8 H 2 + O 2 H 2 O ___ mole O 2 ___ mole H 2 O___ mole H 2 8 507 4.5 6 4 5 3.5 4 3 3 2.5.

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Presentation on theme: "Excess Reactant. Visualizing Limiting Reactant 1 23 4 5 6 7 8 H 2 + O 2 H 2 O ___ mole O 2 ___ mole H 2 O___ mole H 2 8 507 4.5 6 4 5 3.5 4 3 3 2.5."— Presentation transcript:

1 Excess Reactant

2

3 Visualizing Limiting Reactant H 2 + O 2 H 2 O ___ mole O 2 ___ mole H 2 O___ mole H Limiting Reactant Excess Limiting reactant determines amount of product. 2 2

4 Visualizing Limiting Reactant H 2 + O 2 H 2 O ___ mole O 2 ___ mole H 2 O___ mole H Limiting Reactant Excess Limiting reactant determines amount of product. 2 2

5 Moles and Mass Relationships Keys Moles and Mass Relationships Visualizing the Limiting Reactant

6 Visualizing the Limiting Reactants Keys Visualizing the Limiting Reactant

7 Chemistry: Visualizing the Limiting Reactant Use the balanced chemical equation 2 H 2 (g) + O 2 (g) 2 H 2 O(g) for all the problems on this sheet. Directions: Assuming that each molecule shown in the first two containers represents one mole of that substance, write the correct number of moles of substance s below the containers. Then, assume that the contents of the first two containers are combined in the third container. In the third container, draw the correct number of moles of water produced. Key 6 mol H 2 (g) 3 mol O 2 (g) 6 mol H 2 O(g) 1. 8 mol H 2 (g) 4 mol O 2 (g) 8 mol H 2 O(g) 2. Notice: Ratio is the same in both… 2 mol H 2 (g) + 1 mol O 2 (g) 2 mol H 2 O(g)

8 limiting reactant = H 2 excess reactant = O 2 2 mol H 2 (g) 3 mol O 2 (g)2 mol H 2 O(g) + 2 mol O 2 left over limiting reactant = H 2 excess reactant = O 2 6 mol H 2 (g) 4 mol O 2 (g)6 mol H 2 O(g) + 1 mol O 2 left over In the questions above, all the H 2 and O 2 reacted. In most reactions, though, the reactants DO NOT combine perfectly; one reactant will be used up before the other; there is too much of one and not enough of the other. The reactant used up first is called the limiting reactant, the other(s) is/are called the excess reactant(s). Directions cont: For each question below, write the number of moles of substances beneath the corresponding containers. In the third container, draw in the correct number of moles of water produced and any unreacted, excess reactant that is left over. To the right of each question, write the limiting and excess reactant. Key 3. 4.

9 Excess Reactant 2 Na + Cl 2  2 NaCl 50 g 50 g x g “Have” / 23 g/mol/ 71 g/mol 2.17 mol “Need” 1.40 mol EXCESS LIMITING 1.40 mol x 58.5 g/mol 81.9 g NaCl 1 : 2 coefficients 0.70 mol

10 Excess Reactant (continued) 2 Na + Cl 2  2 NaCl 50 g 50 g x g 81.9 g NaCl All the chlorine is used up… 81.9 g NaCl g Cl g Na is consumed in reaction. How much Na is unreacted? 50.0 g g = 18.1 g Na total used “excess” limitingexcess

11 Conservation of Mass is Obeyed 2 Na + Cl 2  2 NaCl 50 g 50 g x g 81.9 g NaCl 2 Na + Cl 2  2 NaCl 50 g x g 81.9 g NaCl 31.9 g g + Na 100 g reactant 100 g product 81.9 g reactant 81.9 g product 50 g

12 excess 125 g Solid aluminum react with chlorine gas to yield solid aluminum chloride. Al(s) + Cl 2 (g) AlCl 3 (s) 322 AlAlCl 3 x g AlCl 3 = 125 g Al 27 g Al = 618 g AlCl 3 1 mol Al 2 mol AlCl 3 2 mol Al x g g AlCl 3 1 mol AlCl 3 Cl 2 AlCl 3 x g AlCl 3 = 125 g Cl 2 71 g Cl 2 = 157 g AlCl 3 1 mol Cl 2 2 mol AlCl 3 3 mol Cl g AlCl 3 1 mol AlCl 3 If 125 g aluminum react with excess chlorine, how many grams of aluminum chloride are made? If 125 g chlorine react with excess aluminum, how many grams of aluminum chloride are made? If 125 g aluminum react with 125 g chlorine, how many grams of aluminum chloride are made? 157 g AlCl 3 We’re out of Cl 2

13 excess 125 g Solid aluminum react with chlorine gas to yield solid aluminum chloride. Al(s) + Cl 2 (g) AlCl 3 (s) 322 AlAlCl 3 x g AlCl 3 = 125 g Al 27 g Al = 618 g AlCl 3 1 mol Al 2 mol AlCl 3 2 mol Al x g g AlCl 3 1 mol AlCl 3 If 125 g aluminum react with excess chlorine, how many grams of aluminum chloride are made? / 27 g/mol 4.6 mol Al Step mol AlCl 3 Step 2 x g/mol 618 g AlCl 3 Step 3 2:2

14 125 g excess Solid aluminum react with chlorine gas to yield solid aluminum chloride. Al(s) + Cl 2 (g) AlCl 3 (s) 322 x g If 125 g chlorine react with excess aluminum, how many grams of aluminum chloride are made? / 71 g/mol Step mol AlCl 3 Step 2 x g/mol 157 g AlCl 3 Step 3 3: mol Cl 2 Cl 2 AlCl 3 x g AlCl 3 = 125 g Cl 2 71 g Cl 2 = 157 g AlCl 3 1 mol Cl 2 2 mol AlCl 3 3 mol Cl g AlCl 3 1 mol AlCl mol Al 2 x mol Al = 3x = 3.52 x = 1.17 mol

15 Limiting Reactants Keys Limiting Reactant Problems

16 1. According to the balanced chemical equation, how many atoms of silver will be produced from combining 100 g of copper with 200 g of silver nitrate? Cu(s) + 2 AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + 2 Ag(s) 2. At STP, what volume of “laughing gas” (dinitrogen monoxide) will be produced from 50 g of nitrogen gas and 75 g of oxygen gas? 3. Carbon monoxide can be combined with hydrogen to produce methanol, CH 3 OH. Methanol is used as an industrial solvent, as a reactant in some synthesis reactions, and as a clean-burning fuel for some racing cars. If you had 152 kg of carbon monoxide and 24.5 kg of hydrogen gas, how many kilograms of methanol could be produced? Limiting Reactant Problems 4. How many grams of water will be produced from 50 g of hydrogen and 100 g of oxygen? Answers: x atoms Ag dm 3 N 2 O kg CH 3 OH g H 2 O Easy

17 1. According to the balanced chemical equation, how many atoms of silver will be produced from combining 100 g of copper with 200 g of silver nitrate? Cu(s) + 2 AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + 2 Ag(s) Limiting Reactant Problems Back 100 g 200 g / 63.5 g/mol / 170 g/mol 1.57 mol Cu 1.18 mol AgNO Limiting Excess x atoms Ag = 1.18 mol AgNO 3 2 mol AgNO 3 2 mol Ag = 7.1 x atoms Ag smaller number is limiting reactant x atoms 1 mol Ag 6.02 x atoms Ag 7.1 x atoms

18 2. At STP, what volume of “laughing gas” (dinitrogen monoxide) will be produced from 50 g of nitrogen gas and 75 g of oxygen gas? Limiting Reactant Problems Back 2 N 2 (g) + O 2 (g) 2 N 2 O(g) 50 g 75 g / 28 g/mol / 32 g/mol 1.79 mol N mol O Limiting Excess x L N 2 O = 1.79 mol N 2 2 mol N 2 2 mol N 2 O = 40 L N 2 O smaller number is limiting reactant x L 1 mol N 2 O 22.4 L N 2 O 40 L N 2 O

19 174.3 g CH 3 OH 32 g CH 3 OH 3. Carbon monoxide can be combined with hydrogen to produce methanol, CH 3 OH. Methanol is used as an industrial solvent, as a reactant in some synthesis reactions, and as a clean-burning fuel for some racing cars. If you had 152 kg of carbon monoxide and 24.5 kg of hydrogen gas, how many kilograms of methanol could be produced? Limiting Reactant Problems Back CO (g) + 2 H 2 (g) CH 3 OH (g) g 24.5 g / 28 g/mol / 2 g/mol 5.45 mol CO mol H Limiting Excess x g CH 3 OH = 5.45 mol CO 1 mol CO 1 mol CH 3 OH = g smaller number is limiting reactant x g 1 mol CH 3 OH Work the entire problem with the mass in grams. At the end, change answer to units of kilograms kg

20 Limiting Reactant Problems 4. How many grams of water will be produced from 50 g of hydrogen and 100 g of oxygen? Back 50 g 100 g / 2 g/mol / 32 g/mol 25 mol H mol O Limiting Excess x g H 2 O = mol O 2 1 mol O 2 2 mol H 2 O = g H 2 O smaller number is limiting reactant x g 1 mol H 2 O 18 g H 2 O g 2 H 2 (g) + O 2 (g) 2 H 2 O(g)

21 5. An unbalanced chemical equation is given as __N 2 H 4 (l) + __N 2 O 4 (l) __N 2 (g) + __ H 2 O(g). If you begin with 400 g of N 2 H 4 and 900 g of N 2 O 4 … Limiting Reactant Problems - continued Easy A. Find the number of liters of water produced (at STP), assuming the reaction goes to completion.at STP 234 B. Find the number of liters of nitrogen produced at STP, assuming the reaction goes to completion. C. Find the mass of excess reactant left over at the conclusion of the reaction. Easy 6. An unbalanced chemical equation is given as __Na(s) + __O 2 (g) __Na 2 O (s) If you have 100 g of sodium and 60 g of oxygen… Easy A. Find the number of moles of sodium oxide produced. B. Find the mass of excess reactant left over at the conclusion of the reaction. Easy Answers: 5A. 560 L H 2 O - gas) 5B. 420 L N 2 5C. 325 g N 2 O 4 excess 6A mol Na 2 O 6B g O 2 excess or 0.45 L H 2 O 42

22 560 L H 2 O 12.5 mol N 2 H 4 2 N 2 H 4 (l) + N 2 O 4 (l) 3 N 2 (g) + 4 H 2 O(g) 5. An unbalanced chemical equation is given as __N 2 H 4 (l) + __N 2 O 4 (l) __N 2 (g) + __ H 2 O(g). If you begin with 400 g of N 2 H 4 and 900 g of N 2 O 4 … Limiting Reactant Problems Back A. Find the number of liters of water produced at STP, assuming the reaction goes to completion. 400 g 900 g / 32 g/mol / 92 g/mol 9.78 mol N 2 O x L H 2 O = 12.5 mol N 2 H 4 2 mol N 2 H 4 4 mol H 2 O = 560 L H 2 O smaller number is limiting reactant x L 1 mol H 2 O 22.4 L H 2 O 234 Water is a SOLID at STP … this isn’t possible! 0.45 L H 2 O x L H 2 O = 12.5 mol N 2 H 4 2 mol N 2 H 4 4 mol H 2 O = 1 mol H 2 O 18 g H 2 O 1.0 g H 2 O 1 mL H 2 O 1000 mL H 2 O 1 L H 2 O Density of water is 1.0 g/mL x L

23 0.45 L H 2 O 12.5 mol N 2 H 4 2 N 2 H 4 (l) + N 2 O 4 (l) 3 N 2 (g) + 4 H 2 O(g) 5. An unbalanced chemical equation is given as __N 2 H 4 (l) + __N 2 O 4 (l) __N 2 (g) + __ H 2 O(l). If you begin with 400 g of N 2 H 4 and 900 g of N 2 O 4 … Limiting Reactant Problems Back A. Find the number of liters of water produced, assuming the reaction goes to completion. 400 g 900 g / 32 g/mol / 92 g/mol 9.78 mol N 2 O x L H 2 O = 12.5 mol N 2 H 4 2 mol N 2 H 4 4 mol H 2 O = smaller number is limiting reactant x L 1 mol H 2 O 18 g H 2 O g H 2 O 1 mL H 2 O 1000 mL H 2 O 1 L H 2 O Density of water is 1.0 g/mL

24 2 N 2 H 4 (l) + N 2 O 4 (l) 3 N 2 (g) + 4 H 2 O(g) 420 L N mol N 2 H 4 5. An unbalanced chemical equation is given as __N 2 H 4 (l) + __N 2 O 4 (l) __N 2 (g) + __ H 2 O(g). If you begin with 400 g of N 2 H 4 and 900 g of N 2 O 4 … Limiting Reactant Problems Back B. Find the number of liters of nitrogen produced at STP, assuming the reaction goes to completion. 400 g 900 g / 32 g/mol / 92 g/mol 9.78 mol N 2 O x L N 2 = 12.5 mol N 2 H 4 2 mol N 2 H 4 3 mol N 2 = smaller number is limiting reactant x L 1 mol N L N 2 234

25 325 g N 2 O 4 excess 92 g N 2 O g N 2 O mol N 2 H 4 2 N 2 H 4 (l) + N 2 O 4 (l) N 2 (g) + H 2 O(g) 5. An unbalanced chemical equation is given as __N 2 H 4 (l) + __N 2 O 4 (l) __N 2 (g) + __ H 2 O(g). If you begin with 400 g of N 2 H 4 and 900 g of N 2 O 4 … Limiting Reactant Problems Back C. Find the mass of excess reactant left over at the conclusion of the reaction. 400 g x g / 32 g/mol x 92 g/mol 6.25 mol N 2 O 4 x g N 2 O 4 = 12.5 mol N 2 H 4 2 mol N 2 H 4 1 mol N 2 O 4 = g N 2 O 4 have - needed 575 g

26 2.17 mol Na 2 O Limiting Reactant Problems 6. An unbalanced chemical equation is given as __Na(s) + __O 2 (g) __Na 2 O (s) If you have 100 g of sodium and 60 g of oxygen… A. Find the number of moles of sodium oxide produced. Back 2.17 mol 4.35 mol Na 4 Na(s) + O 2 (g) 2 Na 2 O (s) 100 g 60 g / 23 g/mol / 32 g/mol mol O x mol Na 2 O = 4.35 mol Na 4 mol Na 2 mol Na 2 O = smaller number is limiting reactant x mol 42

27 x g 34.8 g O g Limiting Reactant Problems 6. An unbalanced chemical equation is given as __Na(s) + __O 2 (g) __Na 2 O (s) If you have 100 g of sodium and 60 g of oxygen… B. Find the mass of excess reactant left over at the conclusion of the reaction g O 2 excess 32 g O g O mol Na Back 100 g / 23 g/mol x 32 g/mol mol O 2 x g O 2 = 4.35 mol Na 4 mol Na 1 mol O 2 = 60 g O 2 have - needed 4 Na(s) + O 2 (g) 2 Na 2 O (s) 42

28 Percent Yield calculated on paper measured in lab % yield = actual yield theoretical yield x 100

29 When 45.8 g of K 2 CO 3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K 2 CO g 46.3 g actual yield excess HCl theoretical yield Theoretical yield x g KCl = 45.8 g K 2 CO 3 = 49.4 g KCl 1 mol K 2 CO g K 2 CO 3 2 mol KCl 1 mol K 2 CO g KCl % Yield = Actual Yield Theoretical Yield % Yield % Yield = 93.7% efficient 49.4 g KCl = x g KCl 1 mol KCl 49.4 g ? g KCl+ H 2 O + CO 2 + H 2 CO 3 2 2

30 g x g Percent Yield W 2 + 2X Y Need 500 g of Y …% yield = 80% actual yield % yield = theoretical yield x atoms x 500 g actual yield x 100 % = x g theoretical yield 0.80 x = 500 g 0.80 x = 625 g x atoms W = 625 g Y 1 mol Y 89 g Y1 mol Y 1 mol W x molecules W 2 1 mol W 2 2 atoms W 1 molecule W x atoms W x L X = 625 g Y 1 mol Y 89 g Y1 mol Y 2 mol X22.4 L X 1 mol X = 315 L X 625 g

31 Cartoon courtesy of NearingZero.net

32 Percent Yield Keys Percent Yield

33 (g)(l) NaHCO 3 HCl ClH HCO 3 Na Baking Soda Lab + + H 2 CO 3 baking soda sodium bicarbonatehydrochloric acidsodium chloride table salt H 2 CO 3 H 2 O + CO 2 (g)  heat NaHCO 3 + HCl NaCl + H 2 O + CO 2 5 g excess gas x g theoretical yield ? g actual yield  % yield = actual yield theoretical yield x 100% Print Copy of Lab Print Copy of Lab

34 Baking Soda Lab Keys Baking Soda Lab Power Point

35 Nuts & Bolts Stoichiometry Lab Keys Nuts & Bolts Lab

36 Smore’s Lab Keys Smore’s Lab

37 Reactions of Copper Lab Keys Reactions of Copper and Percent Yield Power Point

38 WorksheetWorksheet – visualizing limiting reactant visualizing limiting reactant Worksheetvisualizing limiting reactant WorksheetWorksheet – moles and mass relationships moles and mass relationships Worksheetmoles and mass relationships KEYS - Stoichiometry ObjectivesObjectives - stoichiometry Objectives Objectives - mole / chemical formula Objectives (general) Outline (general) Outline LabLab - baking soda lab baking soda lab Labbaking soda lab WorksheetWorksheet - careers in chemistry: farming key key Worksheetkey WorksheetWorksheet - careers in chemistry: dentistry key key Worksheetkey WorksheetWorksheet - easy stoichiometry Worksheet Worksheet - energy energy Worksheetenergy WorksheetWorksheet – percent yield limiting reactants percent yield Worksheetpercent yield WorksheetWorksheet - stoichiometry problems 1 2 stoichiometry problems 12 Worksheetstoichiometry problems 12 WorksheetWorksheet - lecture outline Worksheet Worksheet – limiting reactants Worksheet Worksheet - generic Worksheet LabLab – nuts and bolts Lab S’mores activity TextbookTextbook - questions questions Textbookquestions WorksheetWorksheet - vocabulary vocabulary Worksheetvocabulary

39 Resources - Stoichiometry ObjectivesObjectives - stoichiometry Objectives Objectives - mole / chemical formula Objectives LabLab - baking soda lab Lab WorksheetWorksheet - careers in chemistry: farming key Worksheet Worksheet - careers in chemistry: dentistry key key Worksheetkey WorksheetWorksheet - easy stoichiometry Worksheet Worksheet - energy Worksheet Worksheet - generic Worksheet Worksheet - percent yield Worksheet Worksheet - stoichiometry problems 1 Worksheet (general) Outline (general)general Outlinegeneral WorksheetWorksheet - lecture outline Worksheet TextbookTextbook - questions Textbook WorksheetWorksheet - limiting reactants Worksheet Worksheet – moles and mass relationships Worksheet LabLab – nuts and bolts Lab WorksheetWorksheet – visualizing limiting reactant Worksheet S’mores activity Episode 11 Episode 11 – The Mole WorksheetWorksheet - vocabulary Worksheet


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