CHEM 1215 Exam III John III. Gelder November 10, 1999 Name TA's Name Lab Section INSTRUCTIONS: 1. This examination consists of a total of 7 different pages. The last page includes a periodic table and a solubility table. All work should be done in this booklet. 2. PRINT your name, TA's name and your lab section number now in the space at the top of this sheet. DO NOT SEPARATE THESE PAGES. 3. Answer all questions that you can and whenever called for show your work clearly. Your method of solving problems should pattern the approach used in lecture. You do not have to show your work for the multiple choice (if any) or short answer questions. 4. No credit will be awarded if your work is not shown in problems 3, 5-7 and 8. 5. Point values are shown next to the problem number. 6. Budget your time for each of the questions. Some problems may have a low point value yet be very challenging. If you do not recognize the solution to a question quickly, skip it, and return to the question after completing the easier problems. 7. Look through the exam before beginning; plan your work; then begin. 8. R e l a xand do well. Page 2 Page 3 Page 4 Page 5 TOTAL SCORES (29) (34) (23) (14) (100)
CHEM 1215 EXAM III PAGE 2 (9) 1. Write the chemical formula(s) of the product(s) and balance the following reactions. Identify all products phases as either (g)as, (l)iquid, (s)olid or (aq)ueous. a) H 2 SO 4 (aq) Mg(OH) 2 (aq) MgSO 4 (aq) 2H 2 O(l) b) N 2 (g) 2O 2 (g) 2NO 2 (g) c) Al(NO 3 ) 3 (aq) 3NaOH(aq) Al(OH) 3 (s) 3NaNO 3 (aq) (8) 2. Write the balanced ionic and balanced net ionic chemical equations for the reactions a) and c) in Problem 1. (Remember to include the correct charges on all ions and the phase of each species.) 1a) Ionic equation: 2H (aq) SO 4 2 (aq) 2OH (aq) Mg 2 (aq) Mg 2 (aq) SO 4 2 (aq) 2H 2 O(l) Net Ionic equation: 2H (aq) 2OH (aq) 2H 2 O(l) 1c) Ionic equation: Al 3 (aq) 3NO 3 (aq) 3OH (aq) 3Na (aq) 3Na (aq) 3NO 3 (aq) Al(OH) 3 (s) Net Ionic equation: Al 3 (aq) 3OH (aq) Al(OH) 3 (s) (12)3a. Determine the empirical formula of a compound that is 69.8% carbon, 11.7% hydrogen, and 18.6% oxygen by mass. 69.8 g C 1 mol C 12.0 g C = 5.82 mol C 11.7 g H 1 mol H 1.01 g H = 11.6 mol H 18.6 g O 1 mol O 16.00 g = 1.16 mol O 5.82 mol C 1.16 : 11.6 mol H 1.16 : 1.16 mol O 1.16 5.00 C : 10.0 H : 1.0 O = 5 C : 10 H : 1 O C 5 H 10 O b) If the molar mass of this compound is 172 g mol -1, determine the molecular formula. Empirical mass n = molecular mass n = molecular mass Empirical mass = 172 86 = 2 Molecular formula = (C 5 H 10 O) 2 = C 10 H 20 O 2
CHEM 1215 EXAM III PAGE 3 (22)4. Complete the following table Formula M, Molar m, Mass of n, Moles of N, Number of atoms, Mass g mol sample (g) sample (mol) molecules, or formula units SO 3 80 4.13 x 10 2 5.16 x 10 4 3.11 x 10 20 molecules Na 2 C 2 O 4 134 186 1.39 8.37 x 10 23 formula units unknown 95.8 9.41 x 10 2 9.82 5.91 x 10 24 atoms KPtCl 3 (C 2 H 4 ) 369 7.03 x 10 4 1.91 x 10 6 1.15 x 10 18 formula units Provide the symbol of the unknown element Mo (12) 5. Calculate the following, c) the mass, in grams, of a single atom of mercury. 200.6 g 1 mol 1 mol 6.02 x 10 23 atom = 3.33 x g 10 22 atom d) the number of molecules of water in 1.0 mol of the hydrate CuSO 4 5H 2 O. 1.0 mol CuSO 4 5H 2 O 5 mol H 2 O 1 mol CuSO 4 5H 2 O 6.02 x 1023 molecule 1 mol = 3.01 x 10 24 molecules c) the number of CH 4 molecules and the number of hydrogen atoms in 9.22 g of methane, CH 4. 1 mol CH 4 6.02 x 10 23 molecule CH 4 9.22 g CH 4 16 g 1 mol = 3.47 x 10 23 molecules 3.47 x 10 23 molecules 4 atoms H 1 molecule CH 4 = 1.39 x 10 24 H atoms
CHEM 1215 EXAM III PAGE 4 (13) 6. For the reaction P 4 (s) 10Cl 2 (g) 4PCl 5 (g) Calculate how many grams of chlorine that must react to produce 32.5 g of PCl 5. (Assume phosphorus is in excess.) 32.5 g PCl 5 1 mol PCl 5 208 g 10 mol Cl 2 70.9 g Cl 2 1 mol PCl 5 1 mol Cl 2 = 27.7 g Cl 2 (10) 7. Calculate the mass of silver produced when 3.22 g of Zn react with 4.35 g of AgNO 3. Zn(s) 2AgNO 3 (aq) 2Ag(g) Zn(NO 3 ) 2 (aq) 3.22 g Zn 1 mol 65.4 g = 0.049 mol Zn 4.35 g AgNO 1 mol 3 170.0 g = 0.026 mol AgNO 3 (moles Zn) o 0.049 mol (moles AgNO 3 )required 0.098 mol (moles AgNO 3 ) o 0.026 mol Conclusion AgNO 3 limiting, Zn excess 2 mol AgNO 3 0.049 mol Zn 1 mol Zn = 0.098 mol AgNO 3 0.026 mol AgNO 2 mol Ag 107.9 g 3 2 mol AgNO 3 1 mol Ag = 2.81 g Ag
CHEM 1215 EXAM III PAGE 5 (14) 8. Acetonitile, C 2 H 3 N, is an important nonaqueous solvent. The compound reacts with O 2 according to the equation, 4C 2 H 3 N(g) 15O 2 (g) 8CO 2 (g) 6H 2 O(g) 4NO 2 (g) 47.6 g of acetonitile are added to an amount of oxygen. After the reaction occurs 28.4 grams of H 2 O and 48.4 g of NO 2 are produced. Answer each of the following, a) the mol of H 2 O produced? 1 mol H 2 O 28.4 g H 2 O 18 g = 1.58 mol H 2 O b) the mass of CO 2 produced? 1.58 mol H 2 O 8 mol CO 2 6 mol H 2 O 44.0 g CO 2 1 mol CO 2 = 92.7 g CO 2 c) Could O 2 be the limiting reagent or is it in excess in this reaction? Explain. (You may use a calculation to support your answer.) 1.58 mol H 2 O 4 mol C 2 H 3 N 6 mol H 2 O 41.0 g C 2 H 3 N 1 mol C 2 H 3 N = 43.2 g C 2H 3 N Since we began with 47.8 g of C 2 H 3 N and only 43.2 g reacted, the acetonitrile must be in excess, therefore oxygen is the limiting reagent.
CHEM 1215 EXAM III PAGE 6 (5) 9. THIS IS EXTRA CREDIT. DO NOT ATTEMPT UNTIL AFTER YOU ARE THROUGH DOING THE FIRST 9 QUESTIONS. A 2.24 gram sample of an unknown metal reacts with HCl to produce 0.0808 g of H 2 gas. Identify the metal assuming all of it reacts. (Show ALL your work!) X (s) 2HCl(aq) H 2 (g) XCl 2 (aq) 1 mol H 2 0.0808 g H 1 mol X 2 2.02 g 1 mol H 2 = 0.04 mol X 2.24 g X 0.04 mol X = 56 g mol probably Fe
CHEM 1215 EXAM III PAGE 7 1 2 3 4 5 6 7 IA 1 H 1.008 3 Li 6.94 11 IIA 4 Be 9.01 12 Na Mg 22.99 24.30 19 20 K Ca 39.10 40.08 37 Rb Sr 38 85.47 87.62 55 56 Cs Ba 132.9 137.3 87 88 Fr (223) Ra 226.0 IIIA IVA VA VIA VIIA 4.00 5 6 7 8 B C N O F 9 10 Ne 10.81 12.01 14.01 16.00 19.00 20.18 13 14 15 16 17 18 IIIB IVB VB VIB VIIB VIII IB IIB 26.98 28.09 30.97 32.06 35.45 39.95 21 22 23 24 25 26 27 28 29 Sc Ti V Cr Mn Fe Co Ni Cu 30 31 32 33 34 35 36 Zn Ga Ge As Se Br Kr 44.96 47.88 50.94 52.00 54.94 55.85 58.93 58.69 63.55 65.38 69.72 72.59 74.92 78.96 79.90 83.80 39 Y Zr 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 88.91 91.22 92.91 95.94 (98) 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3 57 72 73 74 76 77 78 79 81 82 83 84 85 86 La Periodic Table of the Elements Hf Al Si P S Cl VIIIA 2 Ta W Re 75 Os Ir Pt Au Hg 80 Tl Pb Bi Po At Rn 138.9 178.5 180.9 183.8 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 (209) (210) 89 104 105 106 107 108 109 Ac 227.0 Rf Db Sg Bh Hs Mt (261) (262) (263) (262) (265) (266) He Ar (222) Lanthanides Actinides 58 59 60 61 62 63 64 Ce Pr Nd Pm Sm Eu Gd 65 66 67 68 69 70 71 Tb Dy Ho Er Tm Yb Lu 140.1 140.9 144.2 (145) 150.4 152.0 157.2 158.9 162.5 164.9 167.3 168.9 173.0 175.0 90 91 92 93 94 95 96 Th Pa U Np Pu AmCm 97 98 99 100 101 102 103 Bk Cf Es Fm Md No Lr 232.0 231.0 238.0 237.0 (244) (243) (247) (247) (251) (252) (257) (258) (259) (260) Solubility Table Ion Solubility Exceptions NO 3 ClO 4 soluble soluble none none Cl soluble except Ag, Hg 2 2, *Pb 2 SO 4 2 CO 3 2 PO 4 3 CrO 4 2 soluble except Ca 2, Ba 2, Sr 2, Hg 2, Pb 2, Ag insoluble except Group IA and NH 4 insoluble except Group IA and NH 4 insoluble except Group IA, IIA and NH 4 - OH insoluble except Group IA, *Ca 2, Ba 2, Sr 2 S 2 insoluble except Group IA, IIA and NH 4 Na soluble none NH 4 soluble none K soluble none *slightly soluble
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