Review Questions (Exam II)

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1 Announcements Exam tonight, 7-8:15pm (locations posted on website) Conflict Exam, 5:15-6:30pm (114 Transportation Bldg) No lab this week! Start new material on Thursday (read chapter 10!)

2 Review Questions (Exam II) 1. b 13. c 22. a) ethanol = LR 2. b 14. e mol CO 2 3. e 15. b mol H 2 O 4. e 16. a b) 6.57 L 5. b 17. d c) 12.0g = 12.0g 6. d 18. c 7. e 19. a 8. c 20. c 9. b 10. a 11. c 12. c

3 Fall 2010 Exam II 1. e 9. a 16. a) same 2. b 10. c b) Soln #1 3. d 11. e c) 6.38 M 4. d 12. a d) 45.5 ml 5. c 13. b 17. b) 37.4 g NF 3 6. c 14. c c) 40. g = 40. g 7. a 15. b d) smaller than 8. e

4 1) 3Fe(s) + 4H 2 O(g) Fe 3 O 4 (s) + 4H 2 (g) How many moles of steam must react to produce 1.62 moles of Fe 3 O 4? 1.62 mol Fe 3 O 4 * 4 mol H 2 O = 6.48 moles H 2 O 1 mol Fe 3 O 4 Answer is (e).

5 2) How many of the following statements about chemical reactions is/are false? I. When balancing a chemical equation, all subscripts must be conserved. II. When one coefficient is doubled, the rest of the coefficients in the balanced equation must also be doubled. III. An individual coefficient in a balanced equation is meaningless. IV. The phases in a chemical reaction tell us the nature of the reactants and products. Answer is (b).

6 3) 3NO 2 (g) + H 2 O(l) 2HNO 3 (aq) + NO(g) Suppose 2.75 moles of HNO 3 were produced from 30.0 g of water and a certain amount of nitrogen dioxide. Which of the following statements is true? a) Water is the limiting reactant because 30.0 g of water produces 2.75 mol HNO 3. b) Water is the limiting reactant because it has a smaller coefficient than the nitrogen dioxide. c) Water is no the limiting reactant because it has a smaller coefficient than the nitrogen dioxide. d) Water is no the limiting reactant because more than 2.75 mol HNO 3 could be produced from 30.0 g of water. e) There is not enough information given to answer this question.

7 3) 3NO 2 (g) + H 2 O(l) 2HNO 3 (aq) + NO(g) Suppose 2.75 moles of HNO 3 were produced from 30.0 g of water and a certain amount of nitrogen dioxide. Which of the following statements is true? How many moles is 30.0 g of water? 30.0 g H 2 O * (mol/18.016g) = 1.67 moles H 2 O How many moles of HNO 3 could I make from that? 1.67 moles H 2 O * (2 mol HNO 3 /1 mol H 2 O) = 3.34 moles HNO3 Since you can possibly make 3.34 moles of HNO 3 from 30.0 g water, but the reaction only made 2.75 moles HNO 3, water cannot be limiting! Answer is (d).

8 4) How many moles of potassium ions are in 80.0 ml of a 2.00M potassium sulfide solution? potassium sulfide: K + and S 2- so formula is K 2 S Moles of K 2 S: 80.0 ml * 1L/1000mL * 2.00 moles/l = 0.16 mol K 2 S Moles of K + : 0.16 moles K 2 S * 2 mol K + = 0.32 moles K + 1 mol K 2 S Answer is (d).

9 5) 2NaN 3 (s) 2Na(s) + 3N 2 (g) What mass of NaN 3 must be used to inflate an air bag to 105L at 1.00 atm and 25 C? Moles of N 2 in air bag: PV = nrt; n = PV/RT n = (1.00 atm)(105l) = mol N 2 ( )(25+273) Moles NaN 3 needed to make N 2 : mol N 2 * 2 mol NaN 3 = mol NaN 3 3 mol N 2 Mass of NaN 3 : mol NaN 3 * (65.01g/mol) = 186 g NaN 3 Answer is (c).

10 6) Fe 2 O 3 (s) + X(s) Fe(l) + X 2 O 3 (s) Balance the equation in standard form and determine the sum of the coefficients. Fe 2 O 3 (s) + 2X(s) 2Fe(l) + X 2 O 3 (s) Sum of coefficients is 6. Answer is (c).

11 7) Fe 2 O 3 (s) + 2X(s) 2Fe(l) + X 2 O 3 (s) If 77.7 g of X reacts with excess iron(iii) oxide to produce 1.44 moles of X 2 O 3, what is the identity of X? Need to determine molar mass of X (g/mol) Grams of X: 77.7 g Moles of X: 1.44 moles X 2 O 3 * 2 mol X = 2.88 mol X 1 mol X 2 O 3 So molar mass would be: 77.7 g X = g/mol, which is aluminum 2.88 mol X Answer is (c).

12 8) Consider the equation: 3A + B 2C. The molar mass of A is 40.0 g/mol. Which of the following statements is true when equal masses of A and B are reacted? a) If the molar mass of B is greater than the molar mass of A, then B must determine how much C is produced. b) If the molar mass of B is less than the molar mass of A, then B must determine how much C is produced. c) If the molar mass of B is the same as the molar mass of A, then A and B react in a perfect stoichiometric ratio and both determine how much C is produced. d) If the molar mass of B is greater than the molar mass of A, then A must determine how much C is produced. e) If the molar mass of B is less than the molar mass of A, then A must determine how much C is produced.

13 8) Consider the equation: 3A + B 2C. The molar mass of A is 40.0 g/mol. Which of the following statements is true when equal masses of A and B are reacted? Assume 40 g of A and B, so A is 1 mole. So moles of B needed to react with 1 mole of A: 1 mol A * (1 mol B/3 mol A) = 0.33 mol B needed If the molar mass of B is greater than the molar mass of A: 40g B * (mol/50g) = 0.8 mol B; A is limiting 40g B * (mol/200g) = 0.2 mol B; B is limiting If the molar mass of B is less than the molar mass of A: 40g B * (mol/30g) = 1.33 mol B; A is limiting 40g B * (mol/10g) = 4.0 mol B; A is limiting If the molar mass of B is the same as the molar mass of A: 40g B * (mol/40g) = 1.0 mol B; A is limiting

14 9) Na 2 SiF 6 (s) + Na(s) Si(s) + NaF(s) Balanced equation: Na 2 SiF 6 (s) + 4Na(s) Si(s) + 6NaF(s) If 3.75 moles of NaF were produced, how many moles of Si were also created? 3.75 mol NaF * 1 mol Si = mol Si 6 mol NaF Answer is (a).

15 10) Na 2 SiF 6 (s) + 4Na(s) Si(s) + 6NaF(s) How many grams of Na 2 SiF 6 (MW = g/mol) are required to completely react with mol Na? Use moles of Na and ratios of coefficients to find moles of Na 2 SiF 6 : mol Na * 1 mol Na 2 SiF 6 = mol Na 2 SiF 6 4 mol Na Then use molar mass to find grams: mol Na 2 SiF 6 * (188.07g/mol) = 26.1 g Na 2 SiF 6 Answer is (c).

16 11) Co 2+ Cl - Na + OH - What is in beaker #1? 4 mol Co 2+ and 8 mol Cl - Molecular formula of compound in beaker #1: CoCl 2 What is in beaker #2? 4 mol Na + and 4 mol OH - Molecular formula of compound in beaker #2: NaOH What products would form in this reaction? Co(OH) 2 and NaCl

17 11) Co 2+ Cl - Na + OH - Unbalanced equation for reaction: CoCl 2 (aq) + NaOH(aq) Co(OH) 2 + NaCl Which one is the precipitate? Co(OH) 2 Balanced equation: CoCl 2 (aq) + 2NaOH(aq) Co(OH) 2 (s) + 2NaCl(aq)

18 11) Co 2+ Cl - Na + OH - Balanced equation: CoCl 2 (aq) + 2NaOH(aq) Co(OH) 2 (s) + 2NaCl(aq) Complete ionic equation: Co 2+ (aq) + 2Cl - (aq) + 2Na + (aq) + 2OH - (aq) Co(OH) 2 (s) + 2Na + (aq) + 2Cl - (aq) Answer is (e).

19 12) Co 2+ Cl - Na + OH - Which of the ions are in solution after the given reaction takes place?

20 12) Co 2+ Cl - Na + OH -

21 12) Which of the following ions are in solution after the given reaction takes place? I. Na + II. OH - III. Co 2+ IV. Cl - Co 2+ Cl - Na + OH - I, III, and IV Answer is (a).

22 13) 10.0 ml of a 0.30M sodium chromate solution reacts with 20.0 ml of a 0.20M aluminum bromide solution. A precipitate forms. What are the reactants? Na + and CrO 4 2- so formula is Na 2 CrO 4 Al 3+ and Br - so formula is AlBr 3 What are the products? Na + and Br - so formula is NaBr Al 3+ and CrO 4 2- so formula is Al 2 (CrO 4 ) 3

23 13) 10.0 ml of a 0.30M sodium chromate solution reacts with 20.0 ml of a 0.20M aluminum bromide solution. A precipitate forms. Unbalanced equation: Na 2 CrO 4 (aq) + AlBr 3 (aq) NaBr + Al 2 (CrO 4 ) 3 Balanced equation: 3Na 2 CrO 4 (aq) + 2AlBr 3 (aq) 6NaBr + Al 2 (CrO 4 ) 3 Which one is the precipitate? NaBr is soluble according to rules so Al 2 (CrO 4 ) 3 is the solid formed

24 13) Balanced equation: 3Na 2 CrO 4 (aq) + 2AlBr 3 (aq) 6NaBr(aq) + Al 2 (CrO 4 ) 3 (s) Moles of reactants: 10.0 ml * (L/1000mL) * 0.30 mol/l = mol Na 2 CrO ml * (L/1000mL) * 0.20 mol/l = mol AlBr 3 Limiting reactant: mol Na 2 CrO 4 * 2 mol AlBr 3 = mol AlBr 3 needed 3 mol Na 2 CrO 4 Have mol AlBr 3 (only need 0.002) so Na 2 CrO 4 is limiting reactant

25 13) Balanced equation: 3Na 2 CrO 4 (aq) + 2AlBr 3 (aq) 6NaBr(aq) + Al 2 (CrO 4 ) 3 (s) Moles of product: mol Na 2 CrO 4 * 1 mol Al 2 (CrO 4 ) 3 = mol Al 2 (CrO 4 ) 3 3 mol Na 2 CrO 4 Mass of product: mol Al 2 (CrO 4 ) 3 * g Al 2 (CrO 4 ) 3 /mol = 0.40 g Al 2 (CrO 4 ) 3 Answer is (b).

26 14) 3Na 2 CrO 4 (aq) + 2AlBr 3 (aq) 6NaBr(aq) + Al 2 (CrO 4 ) 3 (s) What is the concentration of sodium ions left in solution after the reaction is complete? Complete ionic equation: 6Na + (aq) + 3CrO 4 2- (aq) + 2Al 3+ (aq) + 6Br - (aq) 6Na + (aq) + 6Br-(aq) + Al 2 (CrO 4 ) 3 (s) Net ionic equation: 2Al 3+ (aq) + 3CrO 4 2- (aq) Al 2 (CrO 4 ) 3 (s) Na+ does not participate in reaction so: moles Na + in solution before reaction = moles Na + in solution after reaction

27 14) mol Na 2 CrO 4 * 2 mol Na + = mol Na + 1 mol Na 2 CrO 4 Final volume: = 30.0 ml = L Molarity = mol Na + = 0.20M Na L Answer is (c).

28 15) Net ionic equation: 2Al 3+ (aq) + 3CrO 4 2- (aq) Al 2 (CrO 4 ) 3 (s) mol Na 2 CrO 4 (limiting) and mol AlBr 3 Moles of Al 3+ before reaction: mol AlBr 3 * 1 mol Al 3+ = mol Al 3+ Moles of CrO 4 2- : 1 mol AlBr mol Na 2 CrO 4 * 1mol CrO 4 2- = mol CrO mol Na 2 CrO 4

29 Moles of Al 3+ used: mol CrO 2-4 * 2 mol Al 3+ = mol Al 3+ used 3 mol CrO 2-4 Moles of Al 3+ left: = mol Al 3+ Concentration: mol Al 3+ = M Al L Answer is (b).

30 16a) You have three magnesium nitrate solutions on a lab table in front of you. All of the solutions came from a ml volumetric flask containing 3.00 M Mg(NO 3 ) 2. How does the concentration of Solution #1 compare to the concentration of Solution #2? Justify your answer. Solution #1 and Solution #2 have the same concentration (3.00 M) since both solutions came from the same flask (there is still the same ratio of moles/l in each solution).

31 16b) Which solution (not including the volumetric flask) contains the greatest number of ions? Provide complete mathematical support in your answer below. Solution #1: ml of 3.00 M Mg(NO 3 ) ml * L/1000mL * 3.00 mol/l * 3 mol ions/1 mol Mg(NO 3 ) 2 = mol ions Solution #2: 50.0 ml of 3.00 M Mg(NO 3 ) ml * L/1000mL * 3.00 mol/l * 3 mol ions/1 mol Mg(NO 3 ) 2 = mol ions Solution #3: 10.0 ml of 3.00 M Mg(NO 3 ) ml * L/1000mL * 3.00 mol/l * 3 mol ions/1 mol Mg(NO 3 ) 2 = mol ions Solution #1 has the greatest number of ions (0.900 mol)

32 16c) What is the new concentration of Solution #2 if 25.0 g of solid magnesium nitrate (MW = g/mol) is added and dissolved? (Assume no volume change.) 25.0 g Mg(NO 3 ) 2 * 1 mol Mg(NO 3 ) 2 = mol Mg(NO 3 ) 2 added g Mg(NO 3 ) = mol total M = mol = 6.38 M Mg(NO 3 ) L

33 16d) What volume of water (in ml) must evaporate from Solution #1 in order to have a concentration of 5.50 M? M 1 V 1 = M 2 V 2 (3.00 M)(0.100 L) = (5.50 M) * V 2 V 2 = L or 54.5 ml ml = 45.5 ml of water must evaporate

34 17a) N 2 (g) + 3F 2 (g) 2NF 3 (g) Assume this reaction takes place in an elastic balloon with an atmospheric pressure of 1.00 atm and a temperature of 25 C. Scenario I: You have a stoichiometric mixture of nitrogen and fluorine gases (neither reactant is limiting). Draw a molecular-level diagram before and after the reaction occurs, including the relative sizes of the balloons.

35 17b) N 2 (g) + 3F 2 (g) 2NF 3 (g) Assume this reaction takes place in an elastic balloon with an atmospheric pressure of 1.00 atm and a temperature of 25 C. Scenario II: You react 10.0 g nitrogen gas with 30.0 g of fluorine gas at constant temperature and pressure. Determine the mass of NF 3 that can be produced from these two reactants. Moles of N 2 : 10.0g * mol/28.02g = mol N 2 Moles of F 2 : 30.0g * mol/38.00g = mol F 2 Limiting reactant: mol N 2 * 3 mol F 2 /1 mol N 2 = mol F 2 needed Only have mol so F 2 is limiting

36 17b) N 2 (g) + 3F 2 (g) 2NF 3 (g) Assume this reaction takes place in an elastic balloon with an atmospheric pressure of 1.00 atm and a temperature of 25 C. Scenario II: You react 10.0 g nitrogen gas with 30.0 g of fluorine gas at constant temperature and pressure. Determine the mass of NF 3 that can be produced from these two reactants. Moles of product: mol F 2 * 2 mol NF 3 /3 mol F 2 = mol NF 3 Mass of product: mol NF 3 * g/mol = 37.4 g NF 3

37 17c) N 2 (g) + 3F 2 (g) 2NF 3 (g) Assume this reaction takes place in an elastic balloon with an atmospheric pressure of 1.00 atm and a temperature of 25 C. Scenario II: You react 10.0 g nitrogen gas with 30.0 g of fluorine gas at constant temperature and pressure. Verify that mass has been conserved for Scenario II within two significant figures. Mass before the reaction: = 40.0 g total Mass of N 2 left over: mol F 2 * 1 mol N 2 /3 mol F 2 = mol N 2 used = mol N 2 left mol N 2 * g/mol = 2.63 g N 2 left Mass after the reaction: 37.4 g NF g N 2 = 40.0 g total

38 17d) N 2 (g) + 3F 2 (g) 2NF 3 (g) Assume this reaction takes place in an elastic balloon with an atmospheric pressure of 1.00 atm and a temperature of 25 C. For Scenario II, the volume of the balloon after the reaction takes place will be the volume of the balloon before the reaction took place. PV = nrt; at constant P and T: n 1 = n 2 V 1 V 2 n 1 (before): mol N mol F 2 = mol total n 2 (after): mol NF mol N 2 = mol total Since there is a direct relationship between n and V, and n has decreased, the volume will also decrease, so the balloon will be SMALLER.

39 Fall 2011 Exam II 1. d 9. e 16. c) Take 48.1 ml of 2.00 M 2. e 10. c NaCl; add water until you 3. b 11. d get 275 ml total soln 4. a 12. c 5. b 13. a 17. a) i) Beaker D 6. b 14. c ii) Beaker B 7. c 15. a iii) Beaker C 8. d iv) Beaker A b) 10.0 g PbCO 3 c) i) Yes; PbCl 2 ii) NO 3 : M K + : M Na + : M

40 Hour Exam II, Fall ) What information can we obtain from a balanced equation such as 2H 2 (g) + O 2 (g) 2H 2 O(g)? Choose the best answer. a) The formula of each reactant and product b) The phase of each reactant and product c) The starting amount of each reactant d) Both a and b are true e) All of the above (a-c) are true

41 Hour Exam II, Fall ) Consider the following unbalanced equation: C 2 H 6 (g) + O 2 (g) CO 2 (g) + H 2 O(g) Balanced equation: C 2 H 6 (g) + 7/2 O 2 (g) 2CO 2 (g) + 3H 2 O(g) Standard form: 2C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O(g) How many of the following represent a correct mole ratio between C 2 H 6 and O 2 when the equation is balanced? I. 1 : 7/2 II. 2 : 7 III. 4 : 14 IV. 8 : 28 Answer is (e).

42 Hour Exam II, Fall )Fe 2 O 3 (s) + 2Al(s) 2Fe(l) + Al 2 O 3 (s) What mass of iron(iii) oxide must be used to produce 15.0 g of iron? 15.0 g Fe * mol/55.85g = mol Fe mol Fe * 1 mol Fe 2 O 3 = mol Fe 2 O 3 2 mol Fe mol Fe 2 O 3 * g/mol = 21.4 g Fe 2 O 3 Answer is (b).

43 Hour Exam II, Fall )Fe 2 O 3 (s) + 2Al(s) 2Fe(l) + Al 2 O 3 (s) When 15.0 g of iron is produced, what is the maximum mass of aluminum oxide created? mol Fe * 1 mol Al 2 O 3 = mol Al 2 O 3 2 mol Fe mol Al 2 O 3 * g/mol = 13.7 g Al 2 O 3 Answer is (a).

44 Hour Exam II, Fall ) C(s) + O 2 (g) CO 2 (g) What volume of oxygen gas at 30. C and 1.20 atm would be required to react completely with 1.00 g of carbon? 1.00 g C * mol/12.01g = mol C mol C * 1 mol O 2 = mol O 2 PV = nrt; V = nrt/p V = (0.0834)( )(303) = 1.73 L 1.20 Answer is (b).

45 Hour Exam II, Fall ) In which of the following situations would a chemical reaction not take place when the two solutions are mixed? a) AgNO 3 (aq) is mixed with NH 4 Cl(aq) AgCl and NH 4 NO 3 ; AgCl is a solid b) Na 2 CO 3 (aq) is mixed with K 2 S(aq) Na 2 S and K 2 CO 3 ; both are soluble, no net ionic equation c) Ba(NO 3 ) 2 (aq) is mixed with Na 2 SO 4 (aq) BaSO 4 and NaNO 3 ; BaSO4 is a solid d) HCl(aq) is mixed with NaOH(aq) NaCl and H 2 O; no solid, but can write a net ionic equation! e) a chemical reaction would take place in all of the situations above

46 Hour Exam II, Fall ) Unbalanced equation: NH 3 + O 2 NO + H 2 O For every 1.00 mol of NH 3 that reacts, mol of O 2 is required. Balanced equation: 2NH 3 + 5/2 O 2 2NO + 3H 2 O Standard form: 4NH 3 + 5O 2 4NO + 3H 2 O For 1.00 mol NH 3 : 1.00 mol NH 3 * 5 mol O 2 = 1.25 mol O 2 required 4 mol NH 3 Answer is (c).

47 Hour Exam II, Fall ) A g sample of magnesium chloride is measured and weighed. The solid sample is dissolved in water. Enough water is added to make a ml solution. What is the concentration of magnesium chloride in the solution? g MgCl 2 * mol/95.21 g = mol MgCl mol MgCl 2 = M MgCl L Answer is (d).

48 Hour Exam II, Fall ) How many chloride ions are in the solution? mol MgCl 2 * 2 mol Cl - = mol Cl - 1 mol MgCl mol Cl - * x ions = x Cl - ions 1 mol Answer is (e).

49 Hour Exam II, Fall ) Consider the following representation of a chemical equation: Balance the equation in standard form and determine the sum of the coefficients.

50 Hour Exam II, Fall ) Consider the following representation of a chemical equation: = 5 Answer is (c).

51 Hour Exam II, Fall ) Consider the equation: 4A + 3X 2 2A 2 X 3. Which of the following must be conserved? I. Mass II. Moles III. Subscripts IV. Number of atoms Answer is (d).

52 Hour Exam II, Fall ) You add ml of water to ml of a 0.50 M sugar solution. Which of the following will not change? a) total volume of the solution b) concentration of the solution c) moles of solute of the solution d) total mass of the entire solution e) both c and d will not change

53 Hour Exam II, Fall ) 2C 6 H 5 Cl + C 2 HOCl 3 C 14 H 9 Cl 5 + H 2 O Name: chlorobenzene chloral DDT MW: In a government lab, 1150 g of chlorobenzene is reacted with 975 g of chloral. What mass of DDT is formed? 1150 g C 6 H 5 Cl * mol/112.55g = mol C 6 H 5 Cl 975 g chloral * mol/147.38g = 6.62 mol chloral Chloral needed to react: mol C 6 H 5 Cl * 1 mol chloral = 5.11 mol chloral chloral in excess Mass of DDT: 2 mol C 6 H 5 Cl mol C 6 H 5 Cl * 1 mol DDT * g/mol = 1.81 x 10 3 g DDT Answer is (a). 2 mol C 6 H 5 Cl

54 Hour Exam II, Fall ) What mass of excess reactant is left over? Chloral started with: 6.62 mol chloral Chloral needed: mol chloral 1.51 mol chloral left over 1.51 mol chloral * g/mol = 222 g chloral Answer is (c).

55 Hour Exam II, Fall ) As NH 3 (g) is decomposed into nitrogen gas and hydrogen gas at constant pressure and temperature, how does the volume of the product gases collected compare to the volume of NH 3 reacted? Choose the best answer. 2NH 3 (g) N 2 (g) + 3H 2 (g) a) The volume of the product gases is twice the volume of NH 3 reacted because twice the number of moles are produced. b) The volume of the product gases is half the volume of NH 3 reacted because the molecules become smaller in size. c) The volume of the product gases is the same as the volume of NH 3 reacted because the pressure is constant. d) The volume of the product gases is the same as the volume of NH 3 reacted because the law of conservation of mass is followed. e) The volumes cannot be compared to one another without actual data given.

56 Hour Exam II, Fall a) Write and balance the reactions below in standard form, including all phases: (i) Solid magnesium metal reacts with solid manganese(iii) oxide to produce solid magnesium oxide and solid manganese metal. 3Mg(s) + Mn 2 O 3 (s) -> 3MgO(s) + 2Mn(s)

57 Hour Exam II, Fall a) Write and balance the reactions below in standard form, including all phases: (ii) Xenon gas will combine directly with fluorine gas to produce solid xenon tetrafluoride. Xe(g) + 2F 2 (g) -> XeF 4 (s)

58 Hour Exam II, Fall a) Write and balance the reactions below in standard form, including all phases: (iii) Aqueous potassium hydroxide is mixed with aqueous nickel(ii) chloride. Write the balanced molecular, complete ionic, and net ionic equations in standard form and include all phases: Reactants: KOH and NiCl 2 Products: Ni(OH) 2 and KCl

59 Hour Exam II, Fall a) Write and balance the reactions below in standard form, including all phases: (iii) Balanced molecular equation: 2KOH(aq) + NiCl 2 (aq) -> Ni(OH) 2 (s) + 2KCl(aq) Complete ionic equation: 2K + (aq) + 2OH - (aq) + Ni 2+ (aq) + 2Cl - (aq) -> Ni(OH) 2 (s) + 2K + (aq) + 2Cl - (aq) Net ionic equation: 2OH - (aq) + Ni 2+ (aq) -> Ni(OH) 2 (s)

60 Hour Exam II, Fall b) what is a limiting reactant and why is it important in chemistry? Please limit your answer to the space provided. The limiting reactant is consumed (used up) in the reaction. This is important in chemistry because the limiting reactant determines the amount of product that is made.

61 Hour Exam II, Fall c) How do you prepare 275 ml of a M NaCl solution using an available 2.00 M solution? Support your answer with calculations. M 1 V 1 = M 2 V 2 (0.275L)(0.350M) = (x)(2.00m) x = L = 48.1 ml Take 48.1 ml of a 2.00 M NaCl solution and add water until you get 275 ml of total solution (add ml of water).

62 Hour Exam II, Fall a) i) lead(ii) nitrate beaker D. The formula is Pb(NO 3 ) 2, which dissociates to Pb 2+ and 2NO 3-. This represents a 1:2 ratio of a 2+ charge to a 1- charge. ii) sodium chloride beaker B. The formula is NaCl, which dissociates to Na + and Cl -. This represents a 1:1 ratio of a 1+ charge to a 1- charge. iii) potassium carbonate beaker C. The formula is K 2 CO 3, which dissociates to 2K + and CO3 2-. This represents a 2:1 ratio of a 1+ charge to a 2- charge. iv) magnesium sulfate beaker A. The formula is MgSO 4, which dissociates to Mg 2+ and SO This represents a 1:1 ratio of a 2+ charge to a 2- charge.

63 Hour Exam II, Fall b) If 50.0 ml of a 1.00 M lead(ii) nitrate solution is mixed with 75.0 ml of a M potassium carbonate solution, what mass of precipitate will form? Pb(NO 3 ) 2 (aq) + K 2 CO 3 (aq) PbCO 3 (s) + 2KNO 3 (aq) Moles of reactants: 50.0 ml * 1L/1000mL * 1.00 mol Pb(NO 3 ) 2 /L = mol Pb(NO 3 ) ml * 1L/1000mL * mol K 2 CO 3 /L = mol K 2 CO 3 Limiting reactant: mol Pb(NO 3 ) 2 * 1 mol K 2 CO 3 = mol K 2 CO 3 1 mol Pb(NO 3 ) 2 K 2 CO 3 is limiting

64 Hour Exam II, Fall b) If 50.0 ml of a 1.00 M lead(ii) nitrate solution is mixed with 75.0 ml of a M potassium carbonate solution, what mass of precipitate will form? Pb(NO 3 ) 2 (aq) + K 2 CO 3 (aq) PbCO 3 (s) + 2KNO 3 (aq) K 2 CO 3 is limiting Mass of product: mol K 2 CO 3 * 1 mol PbCO 3 * g PbCO 3 = 10.0 g PbCO 3 1 mol K 2 CO 3 mol

65 Hour Exam II, Fall c) After the reaction occurs in part b, 25.0 ml of a 0.500M sodium chloride solution is added. i) Does a precipitate form? Why or why not? Justify your answer. Yes. Pb 2+ is in excess from part b. Pb 2+ reacts with the Cl - to form solid PbCl 2.

66 Hour Exam II, Fall c) After the reaction occurs in part b, 25.0 ml of a 0.500M sodium chloride solution is added. i) What is the concentration of each spectator ion after all of the reactions are complete? (assume the volumes are additive). Spectator ions are K +, NO 3-, and Na +. K + : mol K 2 CO 3 * 2 mol K + = mol K mol = M K L 1 mol K 2 CO 3

67 Hour Exam II, Fall c) After the reaction occurs in part b, 25.0 ml of a 0.500M sodium chloride solution is added. i) What is the concentration of each spectator ion after all of the reactions are complete? (assume the volumes are additive). NO 3- : mol Pb(NO 3 ) 2 * 2 mol NO 3 - = mol NO mol = M NO L 1 mol Pb(NO 3 ) 2

68 Hour Exam II, Fall c) After the reaction occurs in part b, 25.0 ml of a 0.500M sodium chloride solution is added. i) What is the concentration of each spectator ion after all of the reactions are complete? (assume the volumes are additive). Na + : 25.0mL * L/1000mL * mol/l = mol NaCl mol NaCl * 1 mol Na + = mol Na + 1 mol NaCl mol = M Na L

"We are what we repeatedly do. Excellence, therefore, is not an act but a habit." --Aristotle--

CHEMISTRY 101 Hour Exam II October 25, 2016 Adams/Huynh Name Signature Section "We are what we repeatedly do. Excellence, therefore, is not an act but a habit." --Aristotle-- This exam contains 17 questions