Understanding Concepts

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1 UNIT REVIEW (Page 56) Understanding Concepts 1. (a) n NaHCO3 = 3. l M NaHCO3 = 8.01 g/l NaHCO3 = 3.l g NaHCO3 = g =.87 kg The ass of baking soda is.87 kg. (b) n H O =.17 l M H O = 3.0 g/l H O =.17 l 3. 0 g 1ol H O = 73.8 g The ass of hydrogen peroxide is 73.8 g. (c) n Mg(OH) = l M Mg(OH) = g/l Mg(OH) = l g 1ol Mg(OH) = 3.58 g The ass of agnesiu hydroxide is 3.58 g. (d) n CO = l M CO = 8.01 g/l CO = l g CO = g ( 115 g) The ass of carbon noxide is 115 g. (e) n Fe = l (f) M Fe = g/l Fe = l g Fe = g = 66.5 kg The ass of iron is 66.5 kg. n C56 H 88 O = 5.99 l M C56 H 88 O = 793. g/l C56 H 88 O = 5.99 l g 1l C56 H 88 O = g =.75 kg The ass of vitain D is.75 kg.. (a) Ag = 10.5 g M Ag = g/l Copyright 00 Nelson Thoson Learning Unit Quantities in Cheical Reactions 19

2 n Ag = 10.5 g g n Ag = l = 97.3 l The aunt of silver is 97.3 l. (b) C H 5 OH = 8.55 g M C H 5 OH = 6.08 g/l n C H 5 OH = 8.55 g g n C H 5 OH = l ( 186 l) The aunt of ethanol is 186 l. (c) NH NO 3 = g M NH NO = g/l 3 n NH NO = g g n NH NO = 8. l 3 The aunt of aniu nitrate is 8. l. (d) HC H 3 O = 50.0 g M HC H 3 O = g/l n HC H 3 O = 50.0 g g n HC H 3 O = 0.83 l ( 83 l) The aunt of acetic acid is 83 l. (e) C7 H 6 O = 38.9 g (f) M C7 H 6 O = g/l n C7 H 6 O = 38.9 g g n C7 H 6 O = 0.10 ( 101 l) The aunt of cholesterol is 101 l. NaF = g M NaF = 1.99 g/l n NaF = g g n NaF = l = 0.98 l ( 98 µl) The aunt of sodiu fluide is 98 µl. 3. (a) Au = 5.00 g M Au = g/l N A = atos/l n Au = 5.00 g g n Au = 0.05 l N Au = 0.05 l atos l = atos N Au 150 Unit Quantities in Cheical Reactions Copyright 00 Nelson Thoson Learning

3 (b) n Au = l N A = atos/l N Au = l atos l N Au = atos (c) O = 5.00 g M O = 3.00 g/l N A = lecules /l n O = 5.00 g g n O = l N O = l lecules atos 1l 1 lecule N O = atos (d) n O = l N A = lecules/l N O = l lecules atos 1l 1 lecule N O = atos The saple with the st atos is (c): 5.00 g of oxygen gas.. CaCO3 = 335 g M CaCO3 = g/l n CaCO3 = 335 g g n CaCO3 = 3.35 l The aunt of calciu carbonate is 3.35 l. 5. Assue a 100 g saple, f convenience. Pb + = 68.3 g M Pb + = 07.0 g/l S = 10.6 g M S = 3.06 g/l O = 1.1 g M O = g/l n Pb + = 68.3 g g n Pb + = l n S = 10.6 g g n S = 0.33 n O = 1.1 g g = 1.3 l n O The le ratio, Pb + : S : O is : : 1.3. Siplifying (dividing each value by the lowest), we obtain 1.00 : 1.00 :.00, alst exactly 1 : 1 :, aking the epirical fula PbSO (s). Copyright 00 Nelson Thoson Learning Unit Quantities in Cheical Reactions 151

4 6. (a) Ba u u Cr 5.00 u u O u 6.00 u BaCrO u % Ba u 3 u % Ba 5.10% 5.00 u % Cr 100% u % Cr 0.53% 6.00 u % O 100% u % O 5.6% The percentage coposition of BaCrO is 5.10% bariu, 0.53% chroiu, and 5.6% oxygen, by ass. (b) Co u u C 1.01 u u O u u CoCO u u % Co 100% u % Co 9.55% 1.01 u % C 100% u % C 10.10% 8.00 u % O 100% u % O 0.36% The percentage coposition of CoCO 3(s) is 9.55% cobalt, 10.10% carbon, and 0.36% oxygen, by ass. (c) Fe u u Cl 35.5 u u H 1.01 u u O u u FeCl3 6H O 70.3 u u % Fe 100% 70.3 u % Fe 0.66% % Cl u u 15 Unit Quantities in Cheical Reactions Copyright 00 Nelson Thoson Learning

5 % Cl 39.3% 1.1 u % H 100% 70.3 u % H.8% u % O 100% 70.3 u % O 35.51% The percentage coposition of FeCl 3 6H is 0.66% iron, 39.3% chline,.8% hydrogen, and 35.51% oxygen, by ass. 7. (a) Two les of solid nickel(ii) sulfide react with three les of oxygen gas to f two les of solid nickel(ii) oxide and two les of sulfur dioxide gas. The le ratio is :3:: f the reaction as written. (b) Two les of solid aluinu react with three les of aqueous copper(ii) chlide to f two les of aqueous aluinu chlide and three les of solid copper. The le ratio is :3::3 f the reaction as written. (c) Two les of liquid hydrogen peroxide decopose to f two les of liquid water and one le of oxygen gas. The le ratio is ::1 f the reaction as written. 8. (a) NaCl (s) Na (s) + Cl (g) decoposition (b) Na (s) + Na O (s) synthesis (c) Na (s) + H O (l) H (g) + NaOH (aq) single displaceent (d) AlCl 3(aq) + 3 NaOH (aq) Al(OH) 3(s) + 3 NaCl (aq) double displaceent (e) Al (s) + 3 H 3 H (g) + Al (SO ) 3(aq) single displaceent (f) C 8 H 18(l) C + 18 H cobustion 9. (a) 8 Ni (s) + S 8(s) 8 NiS (s) synthesis The le ratio is 8:1:8 as written here. (b) C 6 H 6(l) C + 6 H cobustion The le ratio is :15:1:6 as written here. (c) K (s) + H O (l) KOH (aq) H (g) single displaceent The le ratio is :::1 as written here. 10. Cl (g) KI (aq) I (s) KCl (aq) A diagnostic test f iodine is adding trichlotrifluoethylene and shaking in a closed container. A purple colour in the TTFE layer indicates iodine. Another test would be to do a reverse starch test, i.e., add the ixture to starch to test f iodine. Note: In low concentrations, this reaction produces no precipitate. I (aq) is acceptable as a product on this basis. Copyright 00 Nelson Thoson Learning Unit Quantities in Cheical Reactions 153

6 11. (a) C H (g) 5 C H technological perspective (b) MgCl (s) Mg (s) + Cl (g) scientific perspective (c) Fe (s) 3 Cu 3 Cu (s) Fe (SO ) 3(aq) econoic perspective (d) ZnS (s) 3 ZnO (s) S political perspective (e) Pb(C H 5 ) (l) 7 PbO (s) 16 C 0 H ecological perspective 1. (a) Th He 6 88 Ra (b) 1 8 Pb -1 0 e 1 83 Bi (c) 1 7 N He 1 1 p 17 8 O 13. Assue one le of copound, 16. g. ass % C = 7.0% M C = 1.01 g/l ass % H = 8.7% M H = 1.01 g/l ass % N = 17.3% M N = 1.01 g/l n C = g g n C = 10.0 l 8.7 n H = 16. g g n H = 1 l n N = g g n N =.00 l The integral le ratio is 10 : 1 :, so the lecular fula of nicotine is C 10 H 1 N. 1. (a) In a reagent ix the reagent consued first, causing the reaction to cease, is the liiting reagent. Soe of the other reagent will reain unreacted, so it is said to be in excess. (b) One reagent is present in excess to ensure that all of the other reagent is able to react, and that stoichioetric calculations ade fro that quantity will be accurate. 15. (a) A graph ay be used in quantitative analysis to predict the quantity of one substance involved in the reaction fro the known quantity of another substance. (b) Use of graphs f prediction is very quick and easy, since no calculation is required. 16. The reaction equation coefficients always show the ratio of substances in nuerical aunt (les), so any quantity easureents ust be changed to aunts befe the ratio can be applied. 17. (a) The quantity of product predicted by stoichioetric calculation is the theetical yield. When the reaction is carried out, the easured quantity of product obtained is the actual yield. The percentage yield is the ratio of actual/theetical yields, expressed as a percentage. (b) Yield less than predicted in a reaction ay be due to experiental err inherent in the procedure; to ipurities in the reagents; to unwanted side reactions; and to reactions that are not quantitative that do not go to copletion. 18. Science ephasizes explanation and is international in scope. Technology ephasizes efficient operation ( proble solving) and is usually re local in scope. 15 Unit Quantities in Cheical Reactions Copyright 00 Nelson Thoson Learning

7 19. AgNO 3(aq) Na CrO (aq) NaNO 3(aq) Ag CrO (s) 3.00 g (.81 g actual) g/l g/l n AgNO3 = 3.00 g g = l n AgNO3 n Ag CrO = l 1 n Ag CrO = l Ag CrO = l g 1l Ag CrO =.93 g 1l AgNO 1l Ag Ag CrO = 3.00 g AgNO g A 1l Ag gc C CrO ro 3 ro Ag CrO =.93 g actual yield % yield = 100% the etical yield % yield =. 81 g 100% 95.9%. 93 g The percentage yield of silver chroate in this reaction is 95.9%. 0. (a) BaCl (aq) Na NaCl (aq) BaSO (s) 9.80 g 5.10 g 08.3 g/l 1.0 g/l n BaCl = 9.80 g g n BaCl = 0.07 n Na SO = 5.10 g 1.0 g = l n Na SO Since the reactant le ratio is 1:1, the Na is obviously the liiting reagent f this reaction, and the BaCl (aq) is in excess. (b) The BaCl (aq) is in excess by ( ) l = l. BaCl l g.33 g 1l The BaCl (aq) is in excess by.33 g. Note: The excess of bariu chlide ay also be found by calculating the stoichioetric ass required to react with 5.10 g of sodiu sulfate, and then subtracting that fro 9.80 g. This ethod will produce an answer of.3 g, essentially the sae value, since the third significant digit is uncertain anyway. (c) BaCl (aq) Na NaCl (aq) BaSO (s) 5.10 g 1.0 g/l g/l n Na SO = 5.10 g 1.0 g = l n Na SO n BaSO n BaSO BaSO = l 1 1 = l = l g 1l Copyright 00 Nelson Thoson Learning Unit Quantities in Cheical Reactions 155

8 BaSO = 8.38 g 1l Na BaSO = 5.10 g Na SO 3 SO BaSO g BaSO 1ol B aso = 8.38 g BaSO The ass of bariu sulfate produced would be 8.38 g. 1. (a) 3 Zn (s) CrCl 3(aq) Cr (s) 3 ZnCl (aq) 15.0 g 18.6 g g/l g/l n Zn = 15.0 g g n Zn = 0.9 l = 18.6 g n CrCl3 n CrCl3 = l Since the reactant le ratio is 3:, 0.9 l Zn (s) would require an aunt of 0.9 l /3 = l CrCl 3(aq) f coplete reaction re than is present. The CrCl 3(aq) is obviously the liiting reagent f this reaction, so the Zn (s) is in excess. (b) The l of CrCl 3(aq) present would require an aunt of l 3/ = l Zn (s) f coplete reaction. The Zn (s) is in excess by ( ) l = l. Zn l g 3.5 g The Zn (s) is in excess by 3.5 g. Note: The excess of zinc ay also be found by calculating the stoichioetric ass required to react with 18.6 g of chroiu(iii) chlide, and then subtracting that fro 15.0 g. The sae answer is obtained. (c) 3 Zn (s) CrCl 3(aq) Cr (s) 3 ZnCl (aq) 18.6 g (5.10 g actual) g/l 5.00 g/l n CrCl3 = 18.6 g g = l n CrCl3 n Cr n Cr Cr Cr = l = l = l g 1ol = 6.11 g (theetical) 1l Cl rc Cr = 18.6 g CrCl 3 3 ol Cr g Cr g Cl rc 3 ol Cl rc 3 1ol Cr Cr = 6.11 g (theetical) actual yield % yield = 100% the etical yield % yield = g 100% 83.5% g The percentage yield of chroiu in this reaction is 83.5%.. The hydrated copound ust be FePO XH O (s), where X represents soe nuber of water lecules. The anhydrous copound is FePO (s). The ass of hydrated copound is ( )g 5.13 g The ass of anhydrous copound is (8.7.80)g 3.7 g 156 Unit Quantities in Cheical Reactions Copyright 00 Nelson Thoson Learning

9 The ass of water of hydration is ( )g 1.66 g The lar ass of water, H O, is 18.0 g/l. The lar ass of iron(iii) phosphate, FePO, is g/l. n H O = 1.66 g g n H O = 0.09 n FePO = 3.7 g g = l n FePO The le ratio of FePO :H O is : :.00, so the hydrated copound fula is FePO H O (s). 3. (a) Ag (s) H S (g) Ag S (s) H 0.10 g g/l 7.80 g/l n Ag = 0.10 g g = n Ag n Ag S n Ag S Ag S Ag S Ag S Ag S = = l = l 7.80 g 1l = g ( 138 g) 1l Ag = 0.10 g Ag l Ag S.80 g A g Ag l Ag 17 l o A g gs S = g ( 138 g) The ass of silver sulfide produced would be 138 g. (b) Ag (s) H S (g) Ag S (s) H 0.10 g g/l 3.08 g/l n Ag = 0.10 g g = n Ag n H S n H S H S H S H S H S = = l = l g 1ol = g ( 19.0 g) 1l Ag = 0.10 g Ag l H g Ag l o S Ag g HS 1 ol HS = g ( 19.0 g) The ass of hydrogen sulfide needed would be 19.0 g.. (a) C 1 H O 11(s) H SO (l) 1 C (s) 11 H O (l) H 0.0 g 3.3 g/l 1.01 g/l n C1 H O = 0.0 g g Copyright 00 Nelson Thoson Learning Unit Quantities in Cheical Reactions 157

10 n C1 H O = l 11 n C = l 1 1 n C = 0.70 C = g 1ol = 8. g C C 1 C = 0.0 g C 1 H O 11 1 H O l o 1 ol C1 C H O g C C = 8. g C The ass of carbon fed is 8. g. (b) C 1 H O 11(s) H SO (l) 1 C (s) 11 H O (l) H 0.0 g 3.3 g/l 18.0 g/l n C1 H O = 0.0 g g n C1 H O = l 11 n H O = l n H O = 0.63 l H O = 0.63 l g 1ol = 11.6 g H O C H O = 0.0 g C 1 H O g HO 1 H O 1 H O 11 C 1ol HO 1 H O 11 = 11.6 g H O The ass of water fed is 11.6 g. 5. (a) 6 C 6 H C 6 H 1 O 6(aq) g.01 g/l g/l n C6 H 1 O = g g n C6 H 1 O = l 6 n CO n CO CO CO CO CO = l 6 1 = l = l. 01 g 1ol = 16.6 g 1l C = g C 6 H 1 O 6 6 H 1 O 6l CO. 01 g CO 6 1ol C O = 16.6 g The ass of carbon dioxide required is 16.6 g. (b) 6 C 6 H C 6 H 1 O 6(aq) g g/l 3.00 g/l n C6 H 1 O = g g 158 Unit Quantities in Cheical Reactions Copyright 00 Nelson Thoson Learning

11 n C6 H 1 O 6 = l n O n O O O O = l 6 1 = l = l g 1ol = g 1l C 6 l o O = g C 6 H 1 O g O 6 H 1O g C 1 l o CH 1ol O 6 H O O 6 O = g The ass of oxygen produced is g. 6. NH 3(g) HCl (g) NH Cl (s).00 g.00 g 17.0 g/l 36.6 g/l g/l n NH3 =.00 g g n NH3 = l n HCl =.00 g g = l n HCl The reactant le ratio is 1 : 1, so by inspection HCl (g) is the obvious liiting reagent. n NH Cl n NH Cl NH Cl NH Cl NH Cl NH Cl = l 1 1 = l = l g 1ol =.93 g 1ol Hl C =.00 g HCl 1 l N g l HC 1 l o Cl H Hl C g NHCl 1 ol Nl HC =.93 g The ass of solid aniu chlide produced is.93 g. Applying Inquiry Skills 7. (a) Prediction Ba(OH) (aq) H BaSO (s) H O (l) 3.3 g g/l g/l n Ba(OH) = 3.3 g g n Ba(OH) = l n BaSO n BaSO BaSO BaSO = l 1 1 = l = l g 1l =.67 g Copyright 00 Nelson Thoson Learning Unit Quantities in Cheical Reactions 159

12 1l Ba(OH) 1l BaSO BaSO = 3.3 g Ba(OH) g BaSO 1ol B aso =.67 g BaSO The theetical yield of bariu sulfate is.67 g. To deterine an excess value f H : Ba(OH) (aq) H BaSO (s) H O (l) l g/l g/l n H SO n H SO H SO H SO = l 1 1 = l = l g 1ol = 1.96 g Note: Concentrated sulfuric acid is only about 95% pure, so a larger excess than usual should be used to take this into account. A 0% excess would be: 1.96 g 10% =.35 g (b) Experiental Design A easured ass of bariu hydroxide is dissolved, and reacted with excess sulfuric acid solution. The resulting precipitate is filtered and dried and the ass is then easured. (c) Procedure 1. Sulfuric acid and bariu hydroxide are crosive and soluble bariu copounds are toxic. Use caution in handling and wash your hands after use.. Use a clean, dry 50-L beaker to obtain a 3.3 g saple of Ba(OH) (aq). 3. Use a 100-L beaker to obtain a.35 g saple of concentrated H, and add it to the 50-L beaker.. Allow the precipitate to settle, and test the clear liquid above the precipitate (the supernatant liquid) with a sall aunt of the H fro a edicine dropper to see if further precipitation occurs. 5. If the test in step indicates the reaction is not yet coplete, repeat step until no further precipitation occurs. 6. Measure and recd the ass of a piece of filter paper to 0.01 g. 7. Filter, wash, and dry the BaSO (s) precipitate. 8. Measure and recd the ass of the filter paper plus dry precipitate to 0.01 g. 9. Dispose of all waste aterials accding to instructions. (d) Analysis The ass of precipitate is ( ) g.9 g. (e) Evaluation actual yield % yield = 100% the etical yield % yield =. 9 g. 67 g 100% 91.8% The percentage yield of bariu sulfate in this reaction is 91.8%. A discrepancy of about 8% is considered rather high, so if we assue the design is adequate, either the technique was po the stoichioetric ethod is suspect. 8. The filtrate could be tested with a sall aunt of sulfuric acid. If no precipitate fs, the acid was in excess, and no bariu hydroxide reains. 9. Actual yield can be higher than theetical yield if ipurities exist in the easured product, increasing its ass. Another cause would be not copletely drying a precipitate, which will have the sae effect. 160 Unit Quantities in Cheical Reactions Copyright 00 Nelson Thoson Learning

13 Making Connections 30. Student repts should indicate the postsecondary degrees and courses of study required, and list both industrial and research institutions where analytical cheists ight be eployed. 31. Student repts should include the basic sg reactions. The reaction of nitrogen in internal-cobustion engines: N (g) N The reaction of nitrogen dioxide in sunlight: N (g) ultraviolet light N [O] (atoic oxygen) And the reaction of atoic oxygen to produce ozone: [O] O 3(g) Many further reactions exist as the ozone reacts. The rept should also ephasize that there are two basic ways to reduce the effects of car exhaust in cities: by addressing ileage and engines. The aunt of sg can be reduced by iproving ileage obtained by vehicles, and by reducing ileage, whether by liiting individual trips by increasing the use of public transpt. Sg can also be reduced by lowering nitrogen oxides in exhaust by engine tuning and catalytic conversion. 3. Student repts should explain that sulfuric acid is produced by siply burning sulfur to ake sulfur dioxide and then reacting the S further with to produce SO 3(g). The reaction producing sulfur trioxide does not occur readily, so it is accoplished by ixing the reagents in contact with a hot catalytic (platinu vanadiu pentoxide) surface. The SO 3(g) is reacted with H SO (l) to f H S O 7(l) (pyrosulfuric acid), which is then diluted with water to ref H SO (l). The SO 3(g) is not reacted directly with water because that reaction would release intense heat, and boiling would create a fog stea of hot sulfuric acid droplets a very dangerous situation. The dangerous property of nearly pure (concentrated) sulfuric acid is its aazingly high affinity f water. Contact with this liquid will iediately reve water fro lecules which eans iediate destruction of cells in huan tissue, f exaple. 33. Student repts should explain that aluinu e is a ixture of hydrates of aluinu oxide, Al O 3 xh O, and other oxides like iron(iii) oxide, Fe O 3(s). The ipurities are reved by adding concentrated NaOH (aq), in which the ipurities are not soluble. After filtering out the ipurities, pure aluinu oxide (aluina) is precipitated out. Aluinu is extracted fro lten aluinu oxide at a very high teperature, using electrical energy. Al O 3(l) electricity Al (l) 3 Aluinu selters are only built where large aunts of cheap electricity are available, usually fro a dedicated hydroelectric da and generating station. The teperature required to elt pure aluina is prohibitively high, but can be reduced draatically with the Hall process. This process uses the fact that aluina will dissolve in lten cryolite, Na 3 AlF 6, which elts at about 1000 C, a practical level f industrial wk. GO TO Cheistry 11, Teacher Centre. Expling 3. Student repts f this question will necessarily be personal, the only con criteria being that they describe the question requireents listed: (a) a copany an industry that does this type of wk; (b) the type of analytical techniques used; (c) the level of education and qualifications needed; (d) why this is a career that you ay be interested in. GO TO Cheistry 11, Teacher Centre. Copyright 00 Nelson Thoson Learning Unit Quantities in Cheical Reactions 161

JSUNIL TUTORIAL, SAMASTIPUR

JSUNIL TUTORIAL, SAMASTIPUR Chapter 1 Chemical Reactions and Equations Q 1. Why should a magnesium ribbon be cleaned before burning in air? Ans. Before burning in air, the magnesium ribbon is cleaned by rubbing with a sandpaper.