Unit 3: QUANTITATIVE RELATIONSHIP IN CHEMICAL CHANGE

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1 Unit : Quantitative Relationship in Chemical Change Unit : QUANTITATIVE RELATIONSHIP IN CHEMICAL CHANGE Chapter 8: Chemical Reactions 8.1: Describing Chemical Change Reactants: - chemicals that goes into a reaction. Products: - chemicals that are produced from a reaction. Reactants yields Products Chemical Word Equation: - a chemical reaction written out in words. Skeletal Equation: - a chemical equation that does not show the relative amount of reactants and products. States of Chemicals: - (s) solid, (l) liquid, (g) gas, (aq) aqueous dissolved in water Other Chemical Symbols: 1. Heat is Added: - heat or. Catalyst is Added: - Name of Catalyst (Catalyst: - a chemical that is used to speed up a reaction but does not get consumed) Example 1: Convert the following unbalanced chemical equations into word equations. a. N (g) + H (g) NH (g) Nitrogen gas reacts with hydrogen gas to produce ammonia gas. nitrogen gas + hydrogen gas ammonia gas b. Na (s) + H O (l) NaOH (aq) + H (g) Hydrogen gas and sodium hydroxide solution are produced when solid sodium is added to water. sodium metal + water sodium hydroxide solution + hydrogen gas Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 5.

2 Unit : Quantitative Relationship in Chemical Change Honour Chemistry Example : Convert the following word equations into skeletal equations. a. Heating solid diphosphorous pentaoxide decomposes into phosphorous and oxygen. P O 5 (s) P (s) + O (g) Recall that phosphorus is a polyatomic element, and oxygen is a diatomic element. b. Hypochlorous acid is neutralized by barium hydroxide solution to form water and soluble barium hypochlorite. HClO (aq) + Ba(OH) (aq) H O (l) + Ba(ClO) (aq) It is very important that the subscripts for these ionic compounds are correct. Balancing Chemical Equation: - the process by which we place coefficient (numbers in front of reactants and products) in an attempt to equate the number of atoms or polyatomic ions for the elements or compounds in a chemical equation. Steps involve to Balance Chemical Equation 1. Write the chemical formulas of all reactants and products with their proper subscripts and state to form a skeletal equation.. Count the number of atoms / polyatomic ions (always view complex ion as a group) of each chemical formula. Balance the atoms / polyatomic ions by writing the coefficient in front of the reactant / product. Do NOT mess with any of the subscripts.. Always balance the atoms that appear in more than two chemicals last.. Verify that all atoms / polyatomic ions are balanced. Example : Balance the following chemical equations a. N (g) + H (g) NH (g) b. Al O (s) Al (s) + O (g) N (g) + H (g) NH (g) Al O (s) Al (s) + O (g) c. Na (s) + H O (l) NaOH (aq) + H (g) Na (s) + HOH (l) NaOH (aq) + H (g) Na (s) + HOH (l) NaOH (aq) + H (g) Sometimes, it is easier to rewrite H O as HOH. This is especially true when a OH is part of the products. The first H atom in the HOH becomes H (g) and the remaining part, OH, becomes the polyatomic ion, OH. Page 5. Copyrighted by Gabriel Tang B.Ed., B.Sc.

3 Unit : Quantitative Relationship in Chemical Change d. Fe(NO ) (aq) + Ba(OH) (aq) Fe(OH) (s) + Ba(NO ) (aq) Fe(NO ) (aq) + Ba(OH) (aq) Fe(OH) (s) + Ba(NO ) (aq) Look at each polyatomic ion as one item. There were (NO ) on the left hand side and (NO ) on the right hand side. Hence, we need to use and as coefficients to balance them. Similarly, there were (OH) on the reactant side and (OH) on the product side. Therefore, we are required to use and to balance them. Note that once the coefficients are in place, the Fe and Ba atoms are also balanced. e. C H 6 (g) + O (g) CO (g) + H O (l) For burning (adding O ) with hydrocarbons (compounds containing carbon and hydrogen), we must balance C, H, O in that order. This is because oxygen atoms exist in more than two compounds on the product side. After we balance carbon and hydrogen, we have the following number of oxygen atoms. C H 6 (g) + O (g) CO (g) + H O (l) need 7 oxygen oxygen oxygen C H 6 (g) + 7 O (g) CO (g) + H O (l) Multiply all coefficients by : C H 6 (g) + 7 O (g) CO (g) + 6 H O (l) Example : Rewrite the following word equations into skeletal equations and balance them to form chemical equations. a. Sulfur is burned in the presence of oxygen to form sulfur dioxide gas. Since elemental oxygen is diatomic, the 7 oxygen needed on the reactant side needs to have a coefficient of 7/. This can easily be converted to a whole number by multiplying all the coefficients by. First, we write the skeletal equation. (Recall that sulfur is a polyatomic element, and oxygen is a diatomic element.) Then, we balance the equation. S 8 (s) + O (g) SO (g) S 8 (s) + 8 O (g) 8 SO (g) b. Hydrogen gas is added to propyne (C H ) gas with the aid of a platinum catalyst to form propane gas. First, we write the skeletal equation. (Propane is C H 8 a formula that you should have memorized.) C H (g) + H (g) Then, we balance the equation. Pt C H (g) + H (g) Pt C H 8 (g) C H 8 (g) hydrogen need more hydrogen 8 hydrogen Assignment 8.1 pg. 06 #1, ; pg. 09 #, ; pg. 10 #5 to 7; pg. 11 #8 to 11 Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 55.

4 Unit : Quantitative Relationship in Chemical Change Honour Chemistry 8.: Types of Chemical Reactions There are 5 basic types of chemical reactions: 1. Formation or Composition (Element + Element Compound) Example: Mg (s) + O (g) MgO (s). Deformation or Decomposition (Compound Element + Element) Example: Al O (s) Al (s) + O (g). Single Replacement (Element + Compound Element + Compound) Example: AgNO (aq) + Cu (s) Ag (s) + Cu(NO ) (aq). Double Replacement (Compound + Compound Compound + Compound) Example: AgNO (aq) + NaCl (aq) AgCl (s) + NaNO (aq) 5. Hydrocarbon Combustion (Hydrocarbon + Oxygen Carbon Dioxide + Water) Example: CH (g) + O (g) CO (g) + H O (g) Precipitation Reaction: - a reaction where a precipitate (new solid) is formed as a product. Neutralization Reaction: - a reaction between an acid and a base where water is formed as a product. To Predict Products and Balance Chemical Equations: 1. Write the correct chemical formulas for all products and reactants with proper subscripts. The presence of metals or ionic compounds indicates that we will need to use ions and charges to form any products.. For hydrocarbon combustion, balance in the order of C, H, and then O.. For other type of reactions, balance the equation for each type of cations and anions. Do NOT break up complex ions. Water may be written as HOH (H + and OH ) in single and double replacement reactions.. Check with the Solubility Table (see Section 18.1) and the Table of Elements for the states of chemicals. Example 1: Predict the product(s) along with the states, indicate the type of reaction, and balance the following chemical reactions. a. Sulfur trioxide gas is produced from its elements. Formation: S 8 (s) + 1 O (g) 8 SO (g) b. A solid piece of zinc is immersed in an iron (III) chloride solution. Single Replacement: Zn (s) + FeCl (aq) ZnCl (aq) + Fe (s) Zn + Fe + Cl Page 56. Copyrighted by Gabriel Tang B.Ed., B.Sc.

5 Unit : Quantitative Relationship in Chemical Change c. Propane (C H 8 (g) ) is burned in a gas barbecue. Hydrocarbon Combustion: C H 8 (g) + 5 O (g) CO (g) + H O (g) d. Chlorine gas is bubbled through a copper (II) iodide solution. Single Replacement: Cl (g) + CuI (aq) CuCl (aq) + I (s) Cl Cu + I e. Ammonia gas is decomposed into its elements. Decomposition: NH (g) N (g) + H (g) f. Sulfuric acid is neutralized by sodium hydroxide solution. Double Replacement: H SO (aq) + NaOH (aq) HOH (l) + Na SO (aq) H + SO Na + OH g. Propanol (C H 7 OH (l)) is accidentally ignited. Hydrocarbon Combustion: C H 7 OH (l) + 9 O (g) CO (g) + H O (g) (Multiply Coefficients by ) C H 7 OH (l) + 9 O (g) 6 CO (g) + 8 H O (g) h. Lead (II) nitrate solution is reacted with chromium (III) sulfate solution. Double Replacement: Pb(NO ) (aq) + Cr (SO ) (aq) Cr(NO ) (aq) + PbSO (s) Pb + NO Cr + SO i. Octane (C 8 H 18 (l) ) is combusted in an automobile. Hydrocarbon Combustion: (Multiply Coefficients by ) C 8 H 18 (l) + 5 O (g) 8 CO (g) + 9 H O (g) C 8 H 18 (l) + 5 O (g) 16 CO (g) + 18 H O (g) Assignment 8. pg. 1 #1, 1; pg. 16 #15, 16; pg. 18 #17; pg. 0 1 #18 to 1; pg. #, (omit c) 8.: Reactions in Aqueous Solution Molecular Equation: - a chemical equation where compounds are written in their chemical formulas. Complete Ionic Equation: - a chemical equation where all compounds that are soluble are written in the ionic components (slightly soluble compounds are not separated into ions). Net Ionic Equation: - an ionic equation that only shows the ions responsible in forming the precipitate. Spectator Ions (ions that do not form the precipitate) are omitted. Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 57.

6 Unit : Quantitative Relationship in Chemical Change Honour Chemistry Example 1: Predict all products form when an ammonium phosphate solution reacts with a calcium chloride solution. Explain the reaction in a form of a balanced a. Molecular Equation (NH ) PO (aq) + CaCl (aq) 6 NH Cl (aq) + Ca (PO ) (s) NH + PO Ca + Cl (precipitate) b. Complete Ionic Equation + 6 NH (aq) + PO (aq) + Ca + (aq) + 6 Cl + (aq) 6 NH (aq) + 6 Cl (aq) + Ca (PO ) (s) (Precipitate does NOT separate into ions) c. Net Ionic Equation PO (aq) + Ca + (aq) Ca (PO ) (s) (Only write the ions that contribute to the precipitated chemical species) Example : Predict all products form when sulfuric acid reacts with a lithium hydroxide solution. Explain the reaction in a form of a balanced a. Molecular Equation H SO (aq) + LiOH (aq) Li SO (aq) + HOH (l) H + SO Li + OH (liquid water) b. Complete Ionic Equation Example : Predict all products form when solid aluminum reacts with a copper (II) nitrate solution. Explain the reaction in a form of a balanced a. Molecular Equation H + (aq) + SO (aq) + Li + (aq) + OH (aq) Li + (aq) + SO (aq) + HOH (l) (Pure Liquid does NOT separate into ions) c. Net Ionic Equation H + (aq) + OH (aq) HOH (l) This is the main result of acid-base H + (aq) + OH (aq) H O neutralization (the formation of water). (l) (Only write the ions that contribute to the pure liquid species) Al (s) + Cu(NO ) (aq) Al(NO ) (aq) + Cu (s) Al + Cu + NO b. Complete Ionic Equation (precipitate) Al (s) + Cu + (aq) + 6 NO (aq) Al + (aq) + 6 NO (aq) + Cu (s) c. Net Ionic Equation Al (s) + Cu + (aq) Al + (aq) + Cu (s) (Need to write all the ions on both sides that correspond to any solid used or formed) Page 58. Assignment 8. pg. 6 #5, pg. 8 #6 to 1 Chapter 8 Review: pg. # to 6, 8 to 5 (omit e, 6 & 5c) Copyrighted by Gabriel Tang B.Ed., B.Sc.

7 Unit : Quantitative Relationship in Chemical Change Chapter 9: Stoichiometry 9.1: The Arithmetic of Equations Stoichiometry: - the calculation of quantities in a chemical reaction. - the coefficients of various reactants and /or products form mole ratios. - these mole ratio hold for moles, molecules / atoms, and volumes of gas at constant temperatures and pressures (such as STP and SATP). Example 1: Interpret the chemical equation NH (g) + 7 O (g) NO (g) + 6 H O (g) in terms of a. moles. b. molecules. c. masses. d. volumes STP. e. volumes at SATP. NH (g) 7 O (g) NO (g) 6 H O (g) a. moles of NH 7 moles of O moles of NO 6 moles of H O b. molecules of NH 7 molecules of O molecules of NO 6 molecules of H O c. m = nm m = ( mol)(17.0 g/mol) m = g m = nm m = (7 mol)(.00 g/mol) m =.0 g m = nm m = ( mol)(6.01 g/mol) m = 18.0 g m = nm m = (6 mol)(18.0 g/mol) m = g d. V = ( mol)(. L/mol) V = L V = (7 mol)(. L/mol) V = L V = ( mol)(. L/mol) V = L V = (6 mol)(. L/mol) V = 1. L e. V = ( mol)(.8 L/mol) V = 99.0 L V = (7 mol)(.8 L/mol) V = 17.6 L V = ( mol)(.8 L/mol) V = 99.0 L V = (6 mol)(.8 L/mol) V = 18.8 L Assignment 9.1 pg. 0-1 #,, 6 to 8 Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 59.

8 Unit : Quantitative Relationship in Chemical Change Honour Chemistry 9.A: Chemical Calculations (Gravimetric, STP & SATP Stoichiometry) Mole Ratio: - a ratio form between the coefficient of the required chemical amount to the given chemical require coefficient amount. given coefficient Example 1:.5 mol of PCl 5 (g) is decomposed into its elements. Write a balance equation and determined the amount of chlorine produced. PCl 5 (g) P (s) + 10 Cl (g).5 mol? mol n Cl =.5 mol PCl 5 10 mol Cl mol PCl 5 = mol Cl n Cl = 10.9 mol Gravimetric Stoichiometry: - stoichiometry that involves quantities of masses. Gravimetric Stoichiometry Procedure: 1. Predict the products and balance the chemical equation.. Put all the information given under the appropriate chemicals and determine the molar masses of the chemical involved. m. Find the moles of the given chemical. n = M. Find the mole of the required chemical using mole ratio. require coefficient mol of require = mol of given given coefficient 5. Convert mole of the required chemical to its mass equivalence. (m = nm) Example : Determine the mass of carbon dioxide formed when 50.0 kg of butane (C H 10 (l) ) is burned. C H 10 (g) + 1 O (g) 8 CO (g) + 10 H O (g) 50.0 kg? g 58.1 g/mol M =.01 g/mol C H 10 n = 50.0 kg 58.1 g/mol = kmol C H 10 n CO = kmol C H 10 8 mol CO mol C H 10 =.9978 kmol CO m = nm = (.9978 kmol CO )(.01 g/mol) m CO = 151 kg CO Page 60. Copyrighted by Gabriel Tang B.Ed., B.Sc.

9 Unit : Quantitative Relationship in Chemical Change Example : Barium chloride solution was mixed with an excess sodium phosphate solution. What was the mass of barium chloride solid needed in the original solution to form.1 g of precipitate? BaCl (aq) + Na PO (aq) Ba (PO ) (s) + 6 NaCl (aq)? g.1 g 08. g/mol M = g/mol Ba (PO ) n =.1g g/mol = mol Ba (PO ) n BaCl = mol Ba (PO ) mol BaCl 1mol Ba (PO ) = mol BaCl m = nm = ( mol BaCl )(08. g/mol) m BaCl =. g BaCl Gaseous Stoichiometry (STP and SATP): - stoichiometry that involves quantities of volumes. Gaseous Stoichiometry Procedure (STP and SATP) 1. Predict the products and balance the chemical equation.. Put all the information given under the appropriate chemicals.. Find the moles of the given chemical. 1mol (At STP: n = Volume ; At SATP: n = Volume. L. Find the mole of the required chemical using mole ratio. require coefficient mol of require = mol of given given coefficient 1mol.8 L 5. Convert mole of the required chemical to its volume equivalence. (At STP: V = n. L/mol; At SATP = n.8 L/mol) Example : If 15.5 L of hydrogen at STP is reacted with excess amount of nitrogen, determine the volume of ammonia formed at SATP. ) H (g) + N (g) NH (g) V = 15.5 L? L STP =. L/mol SATP =.8 L/mol n H = 15.5 L 1mol. L n NH = mol H NH = mol mol NH mol H = mol NH V = ( mol NH )(.8 L/mol) = L V NH = 11. L Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 61.

10 Unit : Quantitative Relationship in Chemical Change Honour Chemistry Example 5: A piece aluminum metal is placed in an excess amount of sulfuric acid until all the metal is reacted. Calculate the mass of the aluminum used if 5.76 L of hydrogen gas is evolved at STP. Al (s) + H SO (aq) H (g) + Al (SO ) (aq) M = 6.98 g/mol 5.76 L? g STP =. L/mol n H = 5.76 L 1mol. L = mol mol Al n Al = mol H = mol Al mol H m Al = nm = ( mol Al) (6.98 g/mol) m Al =.6 g Example 6: 5. kg of octane (C 8 H 18 (l) ) is burned under excess oxygen. Determine the volume of carbon dioxide gas produced at SATP. n = C 8 H 18 C 8 H 18 (l) + 5 O (g) 16 CO (g) + 18 H O (g) 5. kg SATP =.8 L/mol M = 11.6 g/mol? L 5. kg = kmol 11.6 g/mol n CO = kmol C 8 H 18 CO 16 mol CO mol C H 8 18 = kmol CO V = ( kmol CO ) (.8 L/mol) V CO = 61. kl Assignment 9.A pg. #9b, 10; pg. 5 #11, 1; pg. 8 #1, 1; pg. 9 #15, 16; pg. 50 #17 to Page 6. Copyrighted by Gabriel Tang B.Ed., B.Sc.

11 Unit : Quantitative Relationship in Chemical Change 9.B: Chemical Calculations (Gaseous and Solution Stochiometry) Gaseous Stoichiometry Procedure (Ideal Gas) 1. Predict the products and balance the chemical equation.. Put all the information given under the appropriate chemicals. PV. Find the moles of the given chemical n =. RT. Find the mole of the required chemical using mole ratio. require coefficient mol of require = mol of given given coefficient 5. Convert mole of the required chemical to its volume equivalence (PV = nrt) Example 1: Ammonia is reacted with oxygen to form nitrogen monoxide and water vapour. n NH = NH (g) + 5 O (g) NO (g) + 6 H O (g) If 50.0 L of ammonia at 90.0 kpa at 5.0 C were allowed to react with excess oxygen, what would be the pressure of nitrogen monoxide in a collector vessel measuring 0.0 L at a temperature of 10.0 C? RT NH (g) + 5 O (g) NO (g) + 6 H O (g) 50.0 L 0.0 L 5.0 C = K 10.0 C = 8.15 K 90.0 kpa? kpa = (90.0 kpa) (50.0 L) PV = mol kpa L ( 8.1 )(98.15K) mol K n NO = mol NH mol NO mol NH = mol NO P NO = nrt V kpa L ( 8 mol)( 8.1 )( 8.15 K) mol K = 0.0 L = 1.56 kpa P NO = 1 kpa Steps to Solve a Precipitation Reaction: 1. Write a balanced molecular equation. Identify the precipitate.. Put the given information underneath the proper chemicals. Identify the limiting reagent if any.. Using n = CV, convert all given information to moles.. Identify and use the information of the limiting reagent if necessary. Require Coefficient 5. Determine the moles of precipitate form by using the mole ratio. Given Coefficient 6. Covert moles of precipitate to mass (m = nm). Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 6.

12 Unit : Quantitative Relationship in Chemical Change Honour Chemistry Filtration: - a separation process to isolate the precipitate formed. Filtration Set-up Folding Filter Paper Note: Wash bottle with distilled water is needed to clean the beaker and the stirring rod. Stirring Rod: - use to guide the unfiltered solution into the filtration apparatus. Precipitate Iron Ring Filter Paper inside Funnel Note: Wash bottle with distilled water is needed to wet the filter paper onto the funnel. Filtrate: - solution that passes through the filter paper. Ring Stand Example : 00 ml of M of calcium chloride is reacted with an excess amount of ammonium phosphate. a. Determine the mass of the precipitate formed in this reaction. b. If the experimental mass of the precipitate is 1.8 g, calculate the % error. How can you interpret this result? a. CaCl (aq) + (NH ) PO (aq) Ca (PO ) (s) + 6 NH Cl (aq) 00 ml = 0.00 L excess? g mol/l M = g/mol n CaCl = CV = ( mol/l)(0.00 L) = 0.01 mol 1mol Ca (PO) n = 0.01 mol CaCl mol CaCl Ca (PO ) = 0.00 mol Ca (PO ) m = nm = (0.00 mol Ca (PO ) )(10.18 g/mol) m Ca (PO ) = 1.0 g Ca (PO ) b. Experimental Theoretiocal 1.8 g 1.0 g % error = 100% = 100% Theoretical 1.0 g Page 6. % error =.% This is a significant error. Since the experimental is much higher than the theoretical, we can say that there were a lot of impurities in the precipitate (from the excess ammonium phosphate). Copyrighted by Gabriel Tang B.Ed., B.Sc.

13 Unit : Quantitative Relationship in Chemical Change Neutralization: - the reaction between acid and base to produce water and salt. Example : HCl (aq) + KOH (aq) HOH (l) + KCl (aq) (Molecular Equation) H + (aq) + Cl (aq) + K + (aq) + OH (aq) HOH (l) + K + (aq) + Cl (aq) (Complete Ionic Equation) H + (aq) + OH (aq) HOH (l) (Net Ionic Equation) Steps to Solve a Neutralization Reaction: 1. Write a balanced molecular equation.. Put the given information underneath the proper chemicals.. Using n = CV, convert the given information to moles. Require Coefficient. Determine the moles of the required chemical by using the mole ratio. Given Coefficient n n 5. Covert moles of the required chemical to concentration or volume C = or V =. V C Titration: - a volumetric analysis that involves measuring the volume of known concentration solution to measure a given volume of an unknown concentration solution. Titration Set-up Titrant: - the solution of known concentration. Buret: - a precise apparatus to deliver the titrant. - the volume of the titrant added is read by subtracting the final volume and the initial volume. Buret Valve: - can be adjusted to let one drop out at a time. Erlenmeyer Flask: - a container commonly uses to hold the analyte. (Narrow mouth prevents splash and spillage.) Analyte: - the solution of an unknown concentration. - the exact volume is usually delivered by a pipet. Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 65.

14 Unit : Quantitative Relationship in Chemical Change Honour Chemistry Acid-Base Titration: - volumetric analysis that assist in determining the unknown concentration in an acid and base neutralization. Equivalent Point (Stoichiometric Point): - a point where the number of moles of H + is equivalent to the number of moles of OH. ( n H + = n OH ) Endpoint: - a point where the indicator actually changes colour to indicate neutralization is completed. Indicator: - a chemical that changes colour due to the ph of the solution. Common Acid-Base Indicators: a. Bromothymol Blue Green at ph = 7 b. Phenol Red Light Orange at ph = 7 c. Phenolphthalein Light Pink at ph = 9 Example : Use the following observation table to determine the concentration of sulfuric acid ml of H SO (aq) titrated by mol/l of KOH (aq) Trial 1 Trial Trial Trial Initial Volume 0. ml.19 ml.8 ml.97 ml Final Volume.19 ml 5.71 ml.97 ml 6.7 ml Volume of KOH added.87 ml 1.5 ml 1.9 ml 1.50 ml Bromothymol Blue Colour Blue Green Green Green First, we have to complete the table by subtracting the final and the initial volumes. Since the titration is completed when the indicator turns green, we only average the result of the last trials. Average Volume of KOH added = 1.5 ml ml ml = 1.50 ml KOH (aq) + H SO (aq) HOH (l) + K SO (aq) 1.50 ml 10.0 ml mol/l? mol/l n KOH = CV = (0.050 mol/l)(1.50 ml) = mmol n = mmol KOH 1mol H SO mol KO H = mmol H SO H SO C H SO = n = V mmol 10.0 ml = mol/l [H SO ] = mol/l Assignment 9.B Gas and Solution Stoichiometry Worksheet Page 66. Copyrighted by Gabriel Tang B.Ed., B.Sc.

15 Unit : Quantitative Relationship in Chemical Change Gas and Solution Stoichiometry Worksheet 1. A 5.0 L propane tank at kpa and 0.0 C is completely used up during a barbecue cookout. Calculate the volume of carbon dioxide when it reached the upper atmosphere where the pressure is at 50.0 kpa and the temperature is at 0.0 C kl of hydrogen at SATP is reacted with excess amount of nitrogen. The resulting ammonia gas needs to be stored in a gas tank with a volume of 6.5 kl and under 000 kpa of pressure. What should the temperature regulator of the gas tank be set at?. Nitrogen and oxygen in the air regularly react within an automobile engine to form nitrogen monoxide. If 0.0 L of oxygen at 00 kpa is reacted inside a 00 C engine, calculate the volume of smog produced at STP in the exhaust.. When nitrogen monoxide reacts with oxygen in the atmosphere under sunlight, it forms photochemical smog (nitrogen dioxide a brown colour gas). If on an average day, 00,000 ML of nitrogen monoxide at SATP is produced in a large city during rush hour; determine the volume of smog produced when it reached a higher altitude with the pressure at 80.0 kpa and the temperature at 15.0 C. 5. A 15.0 g piece of zinc metal is completely reacted with excess amount of hydrochloric acid at 10.0 C and under 90.0 kpa. Determine the volume of gas produced in this reaction ml of 0.50 M of silver nitrate solution is reacted with a large piece of copper wire. Calculate the mass of the precipitate formed in this reaction. 7. Excess aluminum nitrate is reacted with a.6 ml of M strontium hydroxide solution. a. What is the mass of the precipitate formed in this reaction? b. If the experimental mass of the precipitate is 0.9 g, calculate the % error in this experiment. What could contribute to this error? ml of hydrochloric acid is titrated by 0.00 mol/l of sodium hydroxide solution. Determine the concentration of the acid if.8 ml of base is needed to change the bromothymol blue added to green at the endpoint. 9. What is the volume of M of potassium hydroxide needed to completely neutralized 0 ml of 0.5 M phosphoric acid? 10. In an analysis, 15.0 ml of barium hydroxide solution is titrated by mol/l of hypochlorous acid. Calculate the concentration of the base if. ml of acid is required to reach the equivalence point. Answers: L. 9 K. 17 L ML L g 7a g 7b. 8.00%; The experimental value is less than the theoretical. This can mean that some precipitate is lost during filtering or transferring to the funnel mol/l ml mol/l Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 67.

16 Unit : Quantitative Relationship in Chemical Change Honour Chemistry 9.: Calculations Involving Limiting Reagents Excess: - the reactant with more than enough amount for the reaction. Limiting Reagent: - the reactant with the smaller amount (after taken account of the mole ratio) for the reaction. Note: A limiting reagent question will always have enough information to find the moles of both reactants. Steps to deal with Limiting Reagent Problems: 1. Assume one of the reactants is the limiting reagent and determine its mole amount.. Determine the mole amount of the other reactant.. Use the mole amount of the assumed limiting reagent and the mole ratio, calculate the mole amount of the other reactant actually needed.. If the mole amount of the other reactant is smaller than what is needed, then our assumption was wrong. The other reactant is the limiting reagent. 5. If the mole amount of the other reactant is bigger than what is needed, then our assumption was correct. It means that the other reactant is the excess. Example 1: 5.00 g of phosphorus is reacted with g of chlorine gas to produce phosphorus trichloride. Determine the mass of the product produced. P (s) + 6 Cl (g) PCl (s) 5.00 g g? g M = 1.88 g/mol M = g/mol M = 17. g/mol Since there is enough information to determine the moles of two reactants, we need to determine which one is the limiting reagent. n P = m 5.00 g m g = = mol P n Cl = = = mol Cl M 1.88 g/mol M g/mol Let s assume P is the limiting reagent. Calculate the mol Cl actually needed. 6 mol Cl n Cl = mol P = 0.16 mol Cl needed 1mol P But we don t have 0.16 mol of Cl, we only have mol of Cl. Therefore, Cl is the limiting reagent. (Note: the limiting reagent is NOT always the chemical with the smaller number of moles. You have to always compare like we did above.) Now, we calculate the moles of PCl formed by using moles of limiting reagent Cl. mol PCl n PCl = mol Cl = mol PCl 6 mol Cl Finally, we determine the mass of PCl produced. m PCl = nm = ( mol PCl )(17. g/mol) = g m PCl = 19. g Page 68. Copyrighted by Gabriel Tang B.Ed., B.Sc.

17 Unit : Quantitative Relationship in Chemical Change Example : A 0.0 ml of 0.00 M of lead (II) nitrate solution is reacted with 0.0 ml of mol/l of sodium sulfate solution. What is the mass of the precipitate formed at the end of this reaction? Pb(NO ) (aq) + Na SO (aq) PbSO (s) + NaNO (aq) 0.0 ml 0.0 ml? g 0.00 mol/l mol/l M = 0.8 g/mol Since there is enough information to determine the moles of two reactants, we need to determine which one is the limiting reagent. n Pb(NO ) = CV = (0.00 mol/l)(0.0 ml) = 6 mmol n Na SO = CV = (0.150 mol/l)(0.0 ml) = mmol Since the two reactants are of 1:1 ratio, the chemical wth the smaller moles becomes the limiting reagent. (Note: Again, the limiting reagent is NOT always the chemical with the smaller number of moles. We don t have to do the required over given to compare only because they are of 1:1 ratio.) Limiting Reagent is Na SO Now, we calculate the moles of PbSO formed by using moles of limiting reagent Na SO. 1mol PbSO n PbSO = mmol Na SO = mmol PbSO 1mol Na SO Finally, we determine the mass of PbSO produced. m = nm = ( mmol PbSO )(0.8 g/mol) = mg = g PbSO m PbSO = g Assignment 9. pg. 5 #, ; pg. 55 #5; pg. 56 #6; pg. 58 #7, 8; pg. 59 #9 to Chapter 9 Review: pg. 6 # to 1, to 7 Assignment Unit Review: pg. #5 to 58, 60 to 6, 6; pg. 5 #1 to 1 pg. 6 6 #51 to 56, 58; pg. 65 #1 to 1 Copyrighted by Gabriel Tang B.Ed., B.Sc. Page 69.

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