Lecture 5: (Lec3C) Using Stoichiometry

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1 Lecture 5: (Lec3C) Using Stoichiometry Sections (Zumdahl 6 th Edition) Outline: Mass Composition of reactants and products Analyzing Change: The basis of chemical reactions Limiting Reagents in chemical reactions Ch 3 Problems: 3.65, 3.66, 3.68, 3.70, Discussion: 3.4-8, 3.17, 3.18, 3.20, and 3.22

2 Form HF Gas: Macroscopic and Molecular Pictures

3 How to Balance Equations Mass Balance (or Atom Balance)- same number of each element on each side of the equation: (1) start with simplest element (or largest molecule) (2) progress to other elements (3) make all whole numbers (4) re-check each and every atom balance 1 CH 4 (g) + O 2 (g) 1 CO 2 (g) + H 2 O (g) 1 CH 4 (g) + O 2 (g) 1 CO 2 (g) + 2 H 2 O(g) 1 CH 4 (g) + 2 O 2 (g) 1 CO 2 (g) + 2 H 2 O (g) Make charges balance. (Remove spectator ions.) Ca 2+ (aq) + 2 OH - (aq) + Na + Ca(OH) 2 (s) + Na + DEMO: Methane Bubbles

4 Using A Table to Balance Molecular Reaction: Molecular Species ach bo cco dh O CH 4 O 2 CO 2 H 2 O Coefficients a=1 b=2 c=1 d=2 C H O Equations to balance: (Three equations and 4 unknowns) From the first equation: d Second equation: Third equation: 1 a+ 0 b= 1 c+ 0 d 4 a+ 0 b= 0 c+ 2 d 0 a+ 2 b= 2 c+ 1 d a = c= 1 = 2a= 2 b= c+ 1 2 d = 2 3 Equations to Solve Verify the Solution = = =

5 Methane/oxygen reaction CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g) This is the fundamental unit of the reaction, you can t use fewer molecules. All atom types (elements) are conserved All masses are conserved.

6 Information Implied in a Balanced Equation Combustion of Methane Table 3.2 (P 66) Reactants Products CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g) 1 molecule CH molecule CO molecules of O 2 2 molecules H 2 O 1 mol CH 4 molecules + 1 mol CO 2 molecules + 2 mol O 2 molecules 2 mol H 2 O molecules x CH 4 molecules x CO 2 molecules + 2 x (6.022 x ) O 2 molecules 2 x (6.022 x ) H 2 O molecules 16g CH (32g) O 2 44g CO (18g) H 2 O 80g reactants 80g products

7 Information from a balanced reaction

8 Information Contained in a Balanced Equation Viewed in Reactants Products terms of: 2 C 2 H 6 (g) + 7 O 2 (g) = 4 CO 2 (g) + 6 H 2 O (g) + Energy Molecules Amount (mol) Mass (amu) Mass (g) Total Mass (g)

9 Information Contained in a Balanced Equation Viewed in Reactants Products terms of: 2 C 2 H 6 (g) + 7 O 2 (g) = 4 CO 2 (g) + 6 H 2 O (g) + Energy Molecules 2 molecules of C 2 H molecules of O 2 = 4 molecules of CO molecules of H 2 O Amount (mol) 2 mol C 2 H mol O 2 = 4 mol CO mol H 2 O Mass (amu) amu C 2 H amu O 2 = amu CO amu H 2 O Mass (g) g C 2 H g O 2 = g CO g H 2 O Total Mass (g) g = g

10 FIGURE 3.9: Reaction of N 2 + 3H 2 Write the Balanced Reaction for this reaction. How many units of reaction occurred? How many units (i.e. molecules) of H 2 are used per reaction? How many molecules of NH 3 are formed per reaction? If two units of reaction occurred how would it look? Did the reaction go to completion? What is the limiting reactant?

11 Change: The meaning of a chemical reaction A rxn (reaction) represents a change. Burn Sugar: Sugar becomes CO 2 and Water. Sugar on the left (with O 2 ) before reaction Products on the right after the change. Properties of New state must equal properties of old state plus the change between them. For example; We count noses: there are 300 people in this room: Two people leave. How many in the room? We do not need to recount the number of people in the room to know there are 298 people. We kept track of the change, and do the math. New = Initial + Change (NIC Table) Change X Δ X = New Initial

12 Follow the change in the Ammonia Reaction Problem: 0.2 moles N 2 are consumed in the ammonia reaction. How many moles of H 2 are used up and how many moles of NH 3 are produced? 1N + 3H 2 NH The Balanced Reaction 2 2 3

13 A Comment: Stoichiometric Coefficients aa + bb cc + dd ( ) + ( ) ( ) From the reaction 2 moles of NH 3 are produced per mole of reaction. Therefore: 2moles NH = 1mole reaction = 1mol Rxn ( ) ( ) 2 3 moles NH ( ) 3 NH3 sc.. NH3 = c= 2 = 2 mole Rxn Rxn sc.. H = b= 3 General form of a Reaction 1N g 3H g 2 NH g The Balanced Reaction H 2 Rxn The stoichiometric coefficient converts the amount (unit, item, mole, molecule, etc) for one reaction step to the amount for that specific reactant (or product).

14 Follow the change in the Ammonia Reaction Problem: 0.2 moles N 2 are consumed in the ammonia reaction. How many moles of H 2 are used up and how many moles of NH 3 are produced? ( ) ( ) Δ N = New Old = 0 0.2= 0.2moles 2 1N + 3H 2 NH The Balanced Reaction N2 H2 NH SC.. 3 ( N2) = 1 SC..( H2) = 3 SC..( NH3) = 2 Rxn Rxn Rxn Δ ( N2 ) =Δ ( 2 ) = = ( + ) 0.2 = 0.2..( 2 ) SC..( N2 ) = ( ) Δ ( N2 ) = = 0.2 = 0.2. C. ( NH3 ) SC..( N2 ) = ( + ) 0.4 X N moles Rxn moles S C H X moles H X moles Rxn moles S X moles NH 3

15 Balanced reaction! Defines stoichiometric ratios! Unbalanced (i.e., non-stoichiometric) mixture! Limited by syrup!

16 Limiting Reactant in the Ammonia Reaction Problem: 0.2 moles N 2 and 0.40 moles of H 2 are present, initially. The reaction goes to completion, i.e. makes as much ammonia as possible. Which reactant will be consumed first (this is the limiting reactant) and how many moles of NH 3 are produced? 1N + 3H 2 NH The Balanced Reaction 2 2 3

17 Limiting Reactant in the Ammonia Reaction 1N + 3H 2 NH The Balanced Reaction moles N 2 and 0.40 moles of H 2 are present. Determine how much reaction will occur of each reactant independently of the other. 0.20moles N2 x( N2) = 0.20moles N2 = = 0.20moles Rxn SC.. N ( ) ( ) 0.40moles H x H = moles H = = moles Rxn SC..( H2 ) H 2 is the limiting reactant because it produces less reaction. Therefore it is used up completely. Now the problem is equivalent to starting with 0.40 moles H 2 and excess N 2, and computing the amount of ammonia generated. Could also compute the amount of N 2 left when all H 2 is gone. 2

18 Use Balanced reaction to determine mass of product from mass of reactant Typical Problem: Given a chemical reaction. Given the masses of starting material (reactants). Assume: The reaction goes to completion Find: The mass of the product (or one of the products; generally in grams Strategy Balance the chemical reaction (if not balanced) Determine the Molecular masses of needed molecules. Convert starting material from grams to moles. Determine which starting material is the limiting reactant. Assume all of that reactant goes to product. Determine the number of moles of product. Convert the number of moles of product to mass (grams). Done!!

19

20 Sample Problem: Thermite Reaction Problem: Given the following chemical reaction between Aluminum and Iron Oxide. You are asked to make the mixture for the optimum reaction how much Al (in grams) would you add to 1 gram of Iron Oxide? Fe O + 2 Al 2 Fe + Al O The Balanced Reaction

21 Sample Problem: Thermite Reaction Problem: Given the following chemical reaction between Aluminum and Iron Oxide. You are asked to make the mixture for the optimum reaction how much Al (in grams) would you add to 1.00 gram of Iron Oxide? Fe O + 2 Al 2 Fe + Al O The Balanced Reaction MW Fe O g M Al g mol X= Δ= 1.00g FeO = = ( ) mol FeO ( 2 3) = W ( ) = mol ( ) g( Fe2O3) ( ) ( ) MW ( Fe2O3) 1 mol ( Fe2O3) 1 ( ) ( ) SC..( FeO ) X = = + mol Rxn X = mol Rxn S. C. ( Al) = ( ) mol Al X = ( ) mol Al MW ( Al) = ( ) g Al = 0.338g Al

22 Chemical Reactions in Practice: Theoretical, Actual, and Percent Yields Theoretical yield: The amount of product indicated by the stoichiometrically equivalent molar ratio in the balanced equation. Side Reactions: These form smaller amounts of different products that take away from the theoretical yield of the main product. Actual yield: The actual amount of product that is obtained. Percent yield (%Yield): Actual Yield (mass or moles) % Yield = x 100% Theoretical Yield (mass or moles) or Actual yield = Theoretical yield x (% Yield / 100%)

23 Percent Yield What is the actual yield in a chemical reaction if the theoretical yield is 8 grams and the percent yield is 50%. Can the Percent yield be greater than 100% (If yes, give an example.)

24 Calculating the Amounts of Reactants and Products in a Reaction Sequence - I Problem: Calcium Phosphate could be prepared in the following reaction sequence: 3 P + 10 KClO POW!! 4 (s) 3 (s) 3 P 4 O 10 (s) + 10 KCl (s) P 4 O 10 (s) + 6 H 2 O (l) 4 H 3 PO 4 (aq) 2 H 3 PO 4 (aq) + 3 Ca(OH) 2 (aq) 6 H 2 O (aq) + Ca 3 (PO 4 ) 2 (s) Given: 15.5 g P 4 and sufficient KClO 3, H 2 O and Ca(OH) 2. What mass of Calcium Phosphate could be formed? Plan: (1) Calculate moles of P 4. (2) Use molar ratios to get moles of Ca 3 (PO 4 ) 2. (3) Convert the moles of product back into mass by using the molar mass of Calcium Phosphate.

25 Calculating the Amounts of Reactants and Products in a Reaction Sequence - II Solution: moles of P 4 = For Reaction #1 [ 3 P 4 (s) + 10 KClO 3 (s) 3 P 4 O 10 (s) + 10 KCl (s) ] For Reaction #2 [ 1 P 4 O 10 (s) + 6 H 2 O (l) 4 H 3 PO 4 (aq) ] For Reaction #3 [ 2 H 3 PO Ca(OH) 2 1 Ca 3 (PO 4 ) H 2 O] moles Ca 3 (PO 4 ) 2 = moles P 4 x

26 Calculating the Amounts of Reactants and Products in a Reaction Sequence - II Solution: 1 mole P 4 moles of Phosphorous = g P 4 x g P 4 = mol P 4 For Reaction #1 [ 3 P 4 (s) + 10 KClO 3 (s) 3 P 4 O 10 (s) + 10 KCl (s) ] For Reaction #2 [ 1 P 4 O 10 (s) + 6 H 2 O (l) 4 H 3 PO 4 (aq) ] For Reaction #3 [ 2 H 3 PO Ca(OH) 2 1 Ca 3 (PO 4 ) H 2 O] moles P 4 x 3 moles P 4 O 10 x 4 moles H 3 PO 4 x 1 mole Ca 3 (PO 4 ) 2 3 moles P 4 1 mole P 4 O 10 2 moles H 3 PO 4 = moles Ca 3 (PO 4 ) 2

27 Calculating the Amounts of Reactants and Products in a Reaction Sequence - III Molar mass of Ca 3 (PO 4 ) 2 = g mole Mass of Ca 3 (PO 4 ) 2 product =

28 Calculating the Amounts of Reactants and Products in a Reaction Sequence - III Molar mass of Ca 3 (PO 4 ) 2 = g mole g Ca mass of product = moles Ca 3 (PO 4 ) 2 x 3 (PO 4 ) 2 = 1 mole Ca 3 (PO 4 ) 2 = g Ca 3 (PO 4 ) 2

29 Extra Information More Information and extra problems follow. When working extra problems (or problems in the text): Be sure to cover up the solution before you start the problem.

30 Sample Problem: Calculating Reactants and Products in a Chemical Reaction - I Problem: Given the following chemical reaction between Aluminum Sulfide and water, if we are given g of Al 2 S 3 : a) How many moles of water are required for the reaction? b) What mass of H 2 S & Al(OH) 3 would be formed? Al 2 S 3 (s) + 6 H 2 O(l) 2 Al(OH) 3 (s) + 3 H 2 S(g) Plan: Calculate moles of Aluminum Sulfide using its molar mass, then from the equation, calculate the moles of Water, and then the moles of Hydrogen Sulfide, and finally the mass of Hydrogen Sulfide using it s molecular weight. Solution: a) molar mass of Aluminum Sulfide = moles Al 2 S 3 =

31 Sample Problem: Calculating Reactants and Products in a Chemical Reaction - I Problem: Given the following chemical reaction between Aluminum Sulfide and water, if we are given g of Al 2 S 3 : a) How many moles of water are required for the reaction? b) What mass of H 2 S & Al(OH) 3 would be formed? Al 2 S 3 (s) + 6 H 2 O (l) 2 Al(OH) 3 (s) + 3 H 2 S (g) Plan: Calculate moles of Aluminum Sulfide using its molar mass, then from the equation, calculate the moles of Water, and then the moles of Hydrogen Sulfide, and finally the mass of Hydrogen Sulfide using it s molecular weight. Solution: a) molar mass of Aluminum Sulfide = g / mol moles Al 2 S 3 = = moles Al 2 S g Al 2 S g Al 2 S 3 / mol Al 2 S 3

32 Calculating Reactants and Products in a Chemical Reaction - II The ratio of the stoichiometric coefficients gives the ratio of moles used H 2 O: moles Al 2 S 3 x moles H 2 O mole Al 2 S 3 moles H b)h 2 S: moles Al 2 S 3 x 2 S mole Al 2 S 3 molar mass of H 2 S = mass H 2 S = Al(OH) 3 : moles Al 2 S 3 x molar mass of Al(OH) 3 = mass Al(OH) 3 =

33 Calculating Reactants and Products in a Chemical Reaction - II a) cont moles Al 2 S 3 x 6 moles H 2 O = moles H 2 O 1 mole Al 2 S 3 b) moles Al 2 S 3 x 3 moles H 2 S = moles H 2 S 1 mole Al 2 S 3 molar mass of H 2 S = g / mol mass H 2 S = moles H 2 S x g H 2 S = g H 2 S 1 mole H 2 S moles Al 2 S 3 x 2 moles Al(OH) 3 = moles Al(OH) 3 1 mole Al 2 S 3 molar mass of Al(OH) 3 = g / mol mass Al(OH) 3 = moles Al(OH) 3 x g Al(OH) 3 = 1 mole Al(OH) 3 = g Al(OH) 3

34 Molecular model: N 2 molecules require 3H 2 molecules for the reaction N 2 (g) + 3 H 2 (g) 2 NH 3 (g)

35 In the Haber process to make Ammonia You have 9 moles of H 2 reacting with 3 moles of N 2. Assuming the reaction goes to completion, how many moles of each reactant will be remaining?

36 Limiting Reactant Problem: A Typical Problem Problem: A fuel mixture used in the early days of rocketry is composed of two liquids, hydrazine (N 2 H 4 ) and dinitrogen tetraoxide (N 2 O 4 ). They ignite on contact ( hypergolic!) to form nitrogen gas and water vapor. How many grams of nitrogen gas form when exactly 4.50 g N 2 H 4 and 9.00 g N 2 O 4 are mixed? Flow scheme for Problem : Need a balanced reaction, and then: Grams A Moles A Moles B Grams B Extra wrinkle, right now we do not know which is the limiting reagent. We can do the problem assuming Hydrozene is limiting (but we need to figure this out next).

37 A Comment: Stoichiometric Coefficients 2NH l + 1NO l 3N g + 4HO g From the reaction 2 moles of N 2 H 4 liquid are used per mole of reaction. Therefore: 2moles N H = 1mole reaction sc N H ( ) 2 4 ( ) = 2 = 2 sc.. N = 3 ( ) ( ) ( ) ( ) N ( ) moles N H NH mole rxn rxn 2 rxn ( ) 2 4 The stoichiometric coefficient converts the amount (unit, item, mole, molecule, etc) for one reaction step to the amount for that specific reactant (or product).

38 ( ) x= 4.5 g A aka Δx Flow Scheme and Math Calculation ( ) ( ) ( ) ( ) aa + bb cc + dd 2N H l + 1N O l 3N g + 4H O g ( ) mol( A) 1 4.5g A 4.5 x= 4.5 g( A) = = MW ( A) 32 g 32 Moles A mol 4.5 mol ( A) s. c. ( C) 4.5 mol ( A) 3 C mol ( C) x = = rxn = Moles C 32 sc..( A) 32 2 A 2 32 rxn 3 4.5mol ( C) 3 4.5mol ( C) x= MW ( C) = 28 g = 28g ( C) mol 2 32 Grams C x g C g C g A M A sc A sc..( C) = 4.5 ( ) = 5.9 ( ) = ( ) MW ( C) 2 32 ( )..( ) W Grams A At each and every step, X is always X but it is different quantity. Conversion factors and Dimensional Analysis. Reactions are about change: if only 4.5g of 100grams burned to completion, how much N 2 O 4 need be present?

39 Limiting Reactant Problem: A Typical Problem Problem: A fuel mixture used in the early days of rocketry is composed of two liquids, hydrazine (N 2 H 4 ) and dinitrogen tetraoxide (N 2 O 4 ). They ignite on contact ( hypergolic!) to form nitrogen gas and water vapor. How many grams of nitrogen gas form when exactly 4.50 g N 2 H 4 and 9.00 g N 2 O 4 are mixed? Plan: First write the balanced equation. Since amounts of both reactants are given, it is a limiting reactant problem. Calculate the moles of each reactant, and then divide by the equation coefficient to find (the molar equivalent amount) which is limiting and use that one to calculate the moles of nitrogen gas, then calculate mass using the molecular weight of nitrogen gas. Solution: 2NH l + 1NO l 3N g + 4HO g ( ) ( ) ( ) ( ) & Energy Released ( )

40 Hydrazine Rxn Cont d. Use the molecular weights to get the number of moles of reactants M M M W W W ( N ) 2 ( N H ) 2 4 ( N O ) 2 4 = = g mol = = g mol = = g mol ( ) mol( N H ) 4.5g N2H x( NH 2 4) = 4.5g( NH 2 4) = = = 0.14mol g mol 9.0g( N2O4) 9.0mol( N2O4) x( NO 2 4) = 9.0g( NO 2 4) = = = 0.098mol g mol Next Step: Divide by stoichiometric coefficients to see how many moles of reaction (for the reaction as written), to decide which is limiting reagent.

41 Hydrazine Rxn Cont d. ( ) ( ) ( ) M N = M N H = M N O = W 2 W 2 4 W x( N2H4) = mol( N2H4) rxn x( NH 2 4) = 0.14mol( NH 2 4) = 0.070molrxn 2 ( NH) x( N2O4) = mol( N2O4) rxn x NO = 0.098mol NO = 0.098molrxn ( 2 4) ( 2 4) 1 ( NO ) 2 4 g mol Hydrazine is the limiting reagent because it will allow only.07 moles of reaction to occur. This leaves.028 moles N 2 O 4 left over, and we calculated the N 2 yield earlier, 5.90g, assuming hydrazine was limiting.

42 Balancing Chemical Equations - I Problem: The hydrocarbon hexane is a component of Gasoline that burns in an automobile engine to produce carbon dioxide and water as well as energy. Write the balanced chemical equation for the combustion of hexane (C 6 H 14 ). Plan: Write the skeleton equation, converting the words into chemical compounds, with blanks before each compound. Begin the element balance, putting 1 on the most complex compound first, and save oxygen until last! Solution: Balance any elements that this forces:

43 Balancing Chemical Equations - I Problem: The hydrocarbon hexane is a component of Gasoline that burns in an automobile engine to produce carbon dioxide and water as well as energy. Write the balanced chemical equation for the combustion of hexane (C 6 H 14 ). Plan: Write the skeleton equation, converting the words into chemical compounds, with blanks before each compound. Begin the element balance, putting 1 on the most complex compound first, and save oxygen until last! Solution: C 6 H 14 (l) + O 2 (g) CO 2 (g) + H 2 O (g) + Energy Begin with one C 6 H 14 molecule which says that we will get 6 CO 2 s! 1 C 6 H 14 (l) + O 2 (g) 6 CO 2 (g) + H 2 O (g) + Energy

44 Balancing Chemical Equations - II Next balance H atoms: 1 C 6 H 14 (l) + O 2 (g) 6 CO 2 (g) + H 2 O (g) + Energy Balance O atoms last: C 6 H 14 (l) + O 2 (g) CO 2 (g) + H 2 O (g) + Energy C 6 H 14 (l) + O 2 (g) CO 2 (g) + H 2 O (g) + Energy

45 Balancing Chemical Equations - II The H atoms in the hexane will end up as H 2 O, and we have 14 H atoms, and since each water molecule has two H atoms, we will get a total of 7 water molecules. 1 C 6 H 14 (l) + O 2 (g) 6 CO 2 (g) + 7 H 2 O (g) Since oxygen atoms only come as diatomic molecules (two O atoms, O 2 ),we must have even numbers of oxygen atoms on the product side. We do not since we have 7 water molecules! Therefore multiply the hexane by 2, giving a total of 12 CO 2 molecules, and 14 H 2 O molecules. 2 C 6 H 14 (l) + O 2 (g) 12 CO 2 (g) + 14 H 2 O (g) This now gives 12 O 2 from the carbon dioxide, and 14 O atoms from the water, which will be another 7 O 2 molecules for a total of 12+7 =19 O 2! 2 C 6 H 14 (l) + 19 O 2 (g) 12 CO 2 (g) + 14 H 2 O (g)

46 Molecular model: Balanced equation C 2 H 6 O (aq) + 3 O 2(g) 2 CO 2 (g) + 3 H 2 O (g) + Energy

47 Two compounds with the same formula: C 2 H 6 O Structural isomers; with quit different properties; however balanced chemical reactions will look the same.

48 Decomposition of Ammonium Dichromate (NH 4 ) 2 Cr 2 O 7 (s) _ Cr 2 O 3 (s) + _ N 2 (g) + _ H 2 O(g)

49 (NH 4 ) 2 Cr 2 O 7 (s) 1 Cr 2 O 3 (s) + 1 N 2 (g) + 4 H 2 O(g) There are 4 molecular species and 4 atoms. The problem is over-constrained. If it did not balance after balancing the first the atom types, then it cannot be balanced. That didn t happen.

50 Acid - Metal Limiting Reactant - I 2Al (s) + 6HCl (g) 2AlCl 3(s) + 3H 2(g) Consider the reaction above. If we react 30.0 g Al and 20.0 g HCl, how many moles of aluminum chloride will be formed? 30.0 g Al 20.0g HCl Limiting reactant = one w/ fewer equivalents =

51 Acid - Metal Limiting Reactant - I 2Al (s) + 6HCl (g) 2AlCl 3(s) + 3H 2(g) Consider the reaction above. If we react 30.0 g Al and 20.0 g HCl, how many moles of aluminum chloride will be formed? 30.0 g Al / (26.98 g Al / mol Al) = 1.11 mol Al 1.11 mol Al / 2 = equivalents Al 20.0g HCl / (36.5gHCl / mol HCl) = mol HCl O.548 mol HCl / 6 = equivalents HCl HCl is smaller therefore the Limiting reactant!

52 Acid - Metal Limiting Reactant - II Since 6 moles of HCl yield 2 moles of AlCl 3 moles of HCl will yield: x { 2Al(s) + 6HCl(g) 2AlCl 3 (s) + 3H 2 (g) }

53 Acid - Metal Limiting Reactant - II Since 6 moles of HCl yield 2 moles of AlCl moles of HCl will yield: mol HCl x (2 moles of AlCl 3 / 6 mol HCl) = mol of AlCl 3 Therefore there is some Al left over: =0.927 moles

54 What is the quantity of HCl required to react quantitatively with 54 g of aluminum? 1. 1 mol HCl 2. 3 mol HCl 3. 4 mol HCl 4. 6 mol HCl 5. 9 mol HCl 6. None of the above

55 2 Al(s) + 6 HCl(g) 2 AlCl 3 (s) + 3 H 2 (g) Molar mass of Al = 27 g/mol so 54 g Al = 2 mol Al.

56

57 Mg(s) + 2 HCl(g) MgCl 2 (s) + H 2 (g)

58 Limiting Reactant Problems a A + b B + c C d D + e E + f F Steps to solve 1) Identify it as a limiting reactant problem - Information on the: mass, number of moles, number of molecules, volume and molarity of a solution is given for more than one reactant! 2) Calculate moles of each reactant! 3) Divide the moles of each reactant by stoic. coefficient (a,b,c etc...)! 4) Whichever is smallest, that reactant is the limiting reactant! 5) Use the limiting reactant to calculate the moles of product desired then convert to the units needed (moles, mass, volume, number of atoms etc...)!

59 Ostwald Process Limiting Reactant Problem What mass of NO could be formed by the reaction 30.0g of ammonia gas and 40.0g of oxygen gas w/ the rxn below? 4NH 3 (g) + 5 O 2 (g) 4NO (g) + 6 H 2 O (g) 30.0g NH g O 2 fewer, therefore is the Limiting Reagent! Moles NO formed = Mass NO =

60 Ostwald Process Limiting Reactant Problem What mass of NO could be formed by the reaction 30.0g of ammonia gas and 40.0g of oxygen gas in the rxn below? 4NH 3 (g) + 5 O 2 (g) 4NO (g) + 6 H 2 O (g) 30.0g NH 3 / (17.0g NH 3 /mol NH 3 ) = 1.76 mol NH mol NH 3 / 4 = 0.44 moles rxn (eq. NH 3 ) 40.0g O 2 / (32.0g O 2 /mol O 2 ) = 1.25 mol O mol O 2 / 5 = 0.25 eq O 2 Oxygen fewer, therefore oxygen is the Limiting Reagent! 1.25 mol O 2 x 4 mol NO = 1.00 mol NO 5 mol O 2 mass NO = 1.00 mol NO x 30.0 g NO = 30.0 g NO 1 mol NO

61 Flowchart : Solving a stoichiometry problem involving masses of reactants This mole ratio is called the ratio of stoichiometric coefficients

62

63 Percent Yield Problem: Problem: Given the chemical reaction between Iron and water to form the iron oxide, Fe 3 O 4 and hydrogen gas given below. If 4.55 g of iron is reacted with sufficient water to react all of the iron to form rust, what is the percent yield if only 6.02 g of the oxide are formed? Plan: Calculate the theoretical yield and use it to calculate the percent yield, using the actual yield. Solution: 3 Fe (s) + 4 H 2 O (l) Fe 3 O 4 (s) + 4 H 2 (g) Moles Fe = Theoretical moles Fe 3 O 4 = Theoretical mass Fe 3 O 4 = Percent Yield = x 100% = Actual Yield Theoretical Yield

64 Percent Yield Problem: Problem: Given the chemical reaction between Iron and water to form the iron oxide, Fe 3 O 4 and hydrogen gas given below. If 4.55 g of iron is reacted with sufficent water to react all of the iron to form rust, what is the percent yield if only 6.02 g of the oxide are formed? Plan: Calculate the theoretical yield and use it to calculate the percent yield, using the actual yield. Solution: 3 Fe (s) + 4 H 2 O (l) Fe 3 O 4 (s) + 4 H 2 (g) 4.55 g Fe g Fe mol Fe = mol Fe = mol Fe mol Fe x 1 mol Fe 3 O 4 = mol Fe 3 O 3 mol Fe mol Fe 3 O 4 x g Fe 3 O 4 = 6.30 g Fe 3 O 1 mol Fe 4 3 O 4 Actual Yield Percent Yield = x 100% = 6.02 g Fe 3 O 4 x 100% = Theoretical Yield 6.30 g Fe 3 O %

65 Percent Yield / Limiting Reactant Problem - I Problem: Ammonia is produced by the Haber process using nitrogen and hydrogen Gas. If 85.90g of nitrogen are reacted with g hydrogen and the reaction yielded g of ammonia what was the percent yield of the reaction. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Plan: Since we are given the masses of both reactants, this is a limiting reactant problem. First determine which is the limiting reagent then calculate the theoretical yield, and then the percent yield. Solution: moles N 2 = moles H 2 = eqs N 2 = eqs H 2 =

66 Percent Yield / Limiting Reactant Problem - I Problem: Ammonia is produced by the Haber process using nitrogen and hydrogen Gas. If 85.90g of nitrogen are reacted with g hydrogen and the reaction yielded g of ammonia what was the percent yield of the reaction. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Plan: Since we are given the masses of both reactants, this is a limiting reactant problem. First determine which is the limiting reagent then calculate the theoretical yield, and then the percent yield. Solution: Moles of Nitrogen and Hydrogen: moles N 2 = g N 2 = mol N g N mole N 2 moles H 2 = g H 2 = mol H g H mole H 2 Divide by coefficient to get eqvs. of each: g N g H 2 3 = 3.066eq = 3.582eq

67 Percent Yield/Limiting Reactant Problem - II Solution Cont. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) We have moles of Nitrogen, and it is limiting, therefore the theoretical yield of ammonia is: mol NH 3 = mol N 2 x mass NH 3 = Percent Yield = Actual Yield x 100% Theoretical Yield Percent Yield = x 100% = g NH 3 g NH 3

68 Percent Yield/Limiting Reactant Problem - II Solution Cont. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) We have moles of Nitrogen, and it is limiting, therefore the theoretical yield of ammonia is: mol N 2 x 2 mol NH 3 = mol NH 1 mol N 3 2 (Theoretical Yield) g NH mol NH 3 x 3 = g NH 1 mol NH 3(Theoretical 3 Yield) Percent Yield = Actual Yield x 100% Theoretical Yield Percent Yield = x 100% = % g NH g NH 3

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