CHEM 101 HOUR EXAM II 13-OCT-98. Directions: show all work for each question only on its corresponding numbered blue book page.

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1 CHEM 101 HOUR EXAM II 13-OCT-98 Directions: show all work for each question only on its corresponding numbered blue book page. 1. Write a balanced net ionic equation for each reaction or process shown: a. dissolving solid copper(ii) nitrate, Cu(NO 3 ) 2, in water b. adding perchloric acid, HClO 4 - a strong acid, to water c. adding an aqueous solution of acetic acid, HC 2 H 3 O 2 - a weak acid, to an aqueous solution of calcium hydroxide, Ca(OH) 2 - a strong base. d. adding ammonia, NH 3 - a weak base, to water 2. Suppose a mass spectrometer measurement gives the atomic weight of an element as amu. The element is composed of two isotopes; one of mass amu, and the other of amu. What are the abundances of the two isotopes? 3. For the following partially balanced equation: 3 SO 3 + N 2 = 3 SO NH 4 a. without changing any of the coefficients shown, complete the balancing of this equation by adding water and hydrogen ions. b. how many grams of water would be involved if ml of a M Na 2 SO 3 solution were reacted with an excess of nitrogen? 4. Write a balanced net ionic equation for each reaction or process shown: a. mixing an aqueous solution of iron(iii) chloride, FeCl 3, with another aqueous solution containing barium hydroxide, Ba(OH) 2 - a strong base. b. adding an aqueous solution of silver nitrate, AgNO 3, to another aqueous solution with hydrochloric acid, HCl - a strong acid. c. preparing 250 ml of an aqueous solution by dissolving g of potassium permanganate, KMnO 4. d. combining a solution of sodium carbonate, Na 2 CO 3, with a solution of barium nitrate, Ba(NO 3 ) 2.

2 5. For an acid/base titration experiment, ml of M sulfuric acid ( H 2 SO 4 - a strong diprotic acid ) is required to neutralize ml of a potassium hydroxide solution (KOH - a strong base). In this titration, both hydrogen ions from sulfuric acid are neutralized. What is the concentration of the KOH? 6. Balance the following equation which occurs in basic media... SnO 2 + MnO 4 = SnO 3 + Mn 2 O 3 7. For the following partially balanced equation: 3 ClO CrO 2 = 3 ClO + 4 CrO 4 a. without changing any of the coefficients shown, complete the balancing of this equation by adding water and hydroxide ions. b. how many grams of water would be involved if ml of a M NaClO 3 solution were reacted with ml of a M KCrO 2 solution, and the yield was only 90 %? 8. Balance the following equation which occurs in acidic media... PbO 2 + Bi 2 O 3 = PbO 2 + BiO 3 9. a. Write the balanced net ionic equation for the reaction involving the reactants, sodium carbonate, Na 2 CO 3, and uranium(iii) chloride, UCl 3. b. How many mole of precipitate will be formed if g sodium carbonate were reacted with g of uranium(iii) chloride? 10. for the reaction: Ag 2 S + I KCN = 2 I + 2 Ag(CN) 2 + S + 4 K how many grams of potassium cyanide, KCN, are required to form g of sulfur, if the yield for this reaction is only 67 %? Answers to Hour Exam II - 13-OCT-98

3 1. a. Cu(NO 3 ) 2 (s)! Cu 2+ (aq) + 2 NO 3 (aq) complete ionization b. HClO 4 (aq)! H (aq) + ClO 4 (aq) complete ionization c. HC 2 H 3 O 2 (aq) + OH (aq)! HOH + C 2 H 3 O 2 (aq) neutralization d. NH 3 (aq) + HOH = NH 4 (aq) + OH (aq) partial ionization = ( x ) ( y ) x + y = 1, y = 1 - x = ( x ) ( 1 - x ) x = y = a. 3 SO 3 + N H 2 O + 2 H = 3 SO NH 4 C.F. are: THREE SO 3 to THREE H 2 O b. mole H 2 O = mole Na 2 SO 3 [C.F.] [ 3 mole H2O] [ 3mole SO3] = (0.175 M)(0.0250L) = convert mole to grams [ 18 gram]? g = mole H 2 O = g 4. a. Fe 3+ (aq) + 3 OH (aq) = Fe(OH) 3 (s) precipitate formation b. Ag (aq) + Cl (aq) = AgCl (s) precipitate formation c. KMnO 4 (s)! K (aq) + MnO 4 (aq) complete ionization d. Ba 2+ (aq) + CO 3 (aq) = BaCO 3 (s) precipitate formation 5. mole KOH = mole H 2 SO 4 [C.F.] (? M)( L) = (0.101 M)( L)? M KOH = [ 2 mole H ][ 1 mole OH][ 1 mole KOH] [ 1 mole H2SO4][ 1 mole H][ 1 mole OH] 6. a. ox.1/2 cell SnO OH = SnO 3 + H 2 O + 2 e b. red.1/2 cell 2 MnO H 2 O + 6 e = Mn 2 O OH

4 7. a. 3 ClO CrO OH = 3 ClO + 4 CrO H 2 O C.F. s: TWO H 2 O to THREE ClO 3 to FOUR CrO 2 b. a limiting reagent type of problem b/c mole of both reactants are known i. mole H 2 O = mole NaClO 3 [C.F], based on 100 % yield (. )(. ) [ 2 mole H2O 0175M L ] [ 3 mole ClO3] = = ii. iii. mole H 2 O = mole KCrO 2 [C.F.], based on 100 % yield (. )(. ) [ 2 mole H2O 0125M L ] [ 4 mole CrO2] = = limiting reagent is CrO 2. Convert from mole to mass water... mass H 2 O = mole [ 18 gram] = gram iv. However, yield is only 90 %, so mass H 2 O (as product) will be less actual mass H 2 O = ( * 0.9) = g 8. a. red.1/2 cell PbO e = PbO 2 b. ox.1/2 cell Bi 2 O H 2 O = 2 BiO H + 4 e 9. a. 2 U 3+ (aq) + 3 CO 3 (aq) = U 2 (CO 3 ) 3 (s) C.F. s TWO UCl 3 to THREE Na 2 CO 3 to ONE U 2 (CO 3 ) 3 b. this is a limiting reagent type of problem i. mole U 2 (CO 3 ) 3 = mole Na 2 CO 3 [C.F.] ( gram)[ 1 mole ppt] ( 106 g/ mole)[ 3 mole CO3] ii. mole U 2 (CO 3 ) 3 = mole UCl 3 [C.F.] = = ( g)[ 1 mole ppt] ( 656 g/ mole)[ 2 mole U] = = iii. UCl 3 is limiting reagent, and maximum amount of

5 U 2 (CO 3 ) 3 formed w/b moles 10. C.F. s are: FOUR KCN to ONE S a. trial calculation based on 100% yield mole KCN = mole S [C.F.] ( gram)[ 4 mole KCN] ( 32 g/ mole)[ 1 mole S] = = 1.57 mole KCN b. adjust the reactant, KCN, to account for only 67 % yield of product (therefore will need more than 1.57 mole reactant... based on 67 %: mole reactant = 1.57 / 0.67 = mole required c. convert to mass [ 65 gram]? g KCN = mole = g KCN required

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