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KEY SOLUTION TO PRACTICE EXAM II / CHAPTERS 4 & 5 / CHEM 1311 Part I. Multiple Choice (100 Points Total) #01. (5 pts) To which category does the following reaction belong? PbSO4(aq) + K2O(s) PbO(s) + K2SO4(aq) a. a gas-forming reaction b. an acid-base reaction c. a precipitation reaction answer d. an oxidation-reduction reaction e. none of the above #02. (5 pts) Determine the oxidation number of chromium in H 2CrO4. You may want to may use of the oxidation number table on page 8. a. Cr+ b. Cr2+ c. Cr3+ d. Cr4+ e. Cr6+ answer Solution: H = +1; O = -2; Cr = x 2(+1) + x + 4(-2) = 0 2+x8=0 x6=0 x = 6 (i.e. Cr6+) #03. (5 pts) Which one of the following is a weak acid? a. HCl b. HNO3 c. H2SO4 d. HBr e. Acetic acid (C2H4O2) answer (all others are strong acids) #04. (5 pts) Oxidation-reduction reactions always involve ____________ . a. the transfer of electrons b. an oxidizing agent c. a reducing agent d. a precipitate e. a, b & c answer CHEM 1311/TAMUCC/J. M. Southard, PhD 1 #05. (5 pts) Balance the acid-base reaction with the smallest whole number coefficients. I've placed blanks in front of each compound for your convenience. Choose the answer that is the sum of the coefficients in the balanced equation. Do not forget coefficients of "one". 3 Ca(OH)2 + 2 H3PO4(aq) 1 Ca3(PO4)2(aq) a. 4 b. 5 c. 8 d. 9 e. 12 answer (b/c 3 + 2 + 1 + 6 = 12) #06. (5 pts) Based on the following balanced equation determine how many moles of O 2 are required to react with 23.5 moles of CH3OH? 2 CH3OH(l) + 3 O2(g) 2 CO2(g) + 4 H2O(g) a. 47.0 b. 35.3 answer c. 11.8 d. 40.0 e. 23.5 #07. (5 pts) Based on the following balanced equation determine how many moles of H 2 can be produced from 72.9 g Mg and an excess of HCl. 2 HCl(g) + Mg(s) MgCl2(s) + H2(g) a. 4.50 mol b. 3.00 mol answer c. 1.50 mol d. 9.00 mol e. 6.00 mol #08. (5 pts) What is the identity of the "oxidizing agent" in the following oxidation-reduction reaction? 2 Mg(s) + S2(g) 2 MgS(s) a. Mg is "reducing agent b/c Mg Mg2+) b. S2 answer (is "oxidizing agent" b/c S S2-) c. MgS d. This reaction does not involve electron-transfer e. None of the above CHEM 1311/TAMUCC/J. M. Southard, PhD + 6 H2O(l) Solution: 23.5 mol CH3OH 3 mol O2 2 mol CH3OH 35.3 mol O2 Solution: 72.9 g Mg 1 mol Mg 1 mol H 2 24.31 g Mg 1 mol Mg 3.00 mol H 2 2 #09. (5 pts) Based on the following balanced equation, what mass of SiF 4 could be produced by the reaction of 15.0 g SiO2 with an excess of HF? SiO2(s) FW 60.08 + 4 HF(l) FW 20.01 SiF4(l) FW 104.08 + 2 H2O(l) FW 18.02 a. 1.04 g b. 12.0 g c. 26.0 ganswer d. 104 g e. 52.0 g Solution: Note: Since HF is in excess, SiO2 is therefore the limiting reactant. 15.0 g SiO 2 1 mol SiO 2 1 mol SiF4 104.08 g SiF4 1 mol SiF4 60.08 g SiO 2 1 mol SiO 2 26.0 g SiF4 #10. (5 pts) What is the net ionic equation for the following formula unit equation? Cu(NO3)2(aq) + Na2CO3(aq) CuCO3(s) + 2 NaNO3(aq) Total ionic equation (cross out spectator ions for net ionic equation) Cu2+(aq) + 2 NO3-(aq) + 2 Na+(aq) + CO32-(aq) CuCO3(s) + 2 Na+(aq) + 2 NO3-(aq) a. Cu2+(aq) + CO32-(aq) CuCO3(s) answer b. Cu2+(aq) + 2 NO3-(aq) + 2 Na+(aq) + CO32-(aq) CuCO3(s) + 2 Na+(aq) + 2 NO3-(aq) c. Cu2+(aq) + 2 NO3-(aq) + 2 Na+(aq) CuCO3(s) + 2 Na+(aq) + 2 NO3-(aq) d. Cu2+(aq) + CO32-(aq) Cu2+(s) + CO32-(s) e. 2 NO3-(aq) + 2 Na+(aq) 2 NaNO3(s) #11. (5 pts) Based upon the following balanced equation, how many grams of HNO 3 can be prepared from the reaction of 69.0 grams of NO 2 with 36.0 grams of H2O? 3 NO2(g) + FW 46.01 H2O(l) FW 18.02 2 HNO3(aq) + FW 63.01 NO(g) FW 30.01 Notes: We have 2.00 mol H2O We need 0.50 mol H2O So H2O is "in excess" We have 1.50 mol NO2 We need 6.00 mol NO2 So NO2 is the "limiting reactant" a. 252 g b. 63.0 g answer c. 116 g d. 84.0 g e. 76.0 g Moles on hand: 1 mol NO 2 69.0 g NO 2 1.50 mol NO 2 46.01 g NO 2 1 mol H 2O 36.0 g H 2O 2.00 mol H 2O 18.02 g H 2O Moles needed: 1 mol H 2 O 1.50 mol NO 2 0.50 mol H 2 O 3 mol NO 2 CHEM 1311/TAMUCC/J. M. Southard, PhD 3 2.00 mol H 2O 3 mol NO 2 1 mol H 2O 6.00 mol NO 2 1.50 mol NO 2 63.01 g HNO 3 1 mol HNO 3 2 mol HNO 3 3 mol NO 2 63.0 g HNO 3 #12. (5 pts) What is the percent yield of elemental sulfur (S) if 7.54 grams of sulfur are obtained from the reaction of 6.16 grams of SO2 with an excess of H2S? 2 H2S(g) + FW 34.08 SO2(g) FW 64.06 2 H2O(l) + FW 18.02 3 S(s) FW 32.07 a. 72.6% b. 40.8% c. 81.5% answer d. 88.4% e. 91.4% 1 calculate theoretical yield: 6.16 g SO 2 1 mol SO 2 64.06 g SO 2 3 mol S 1 mol SO 2 32.07 g S 1 mol S 9.25 g SO 2 st 2nd calculate percent yield: %Yield 7.54 g 100 9.25 g 81.5% #13. (5 pts) You produce aspirin in lab. Theoretically, you determine that 6.25 grams aspirin may be produced. However, after you isolate your aspirin you determine that made 4.82 grams of aspirin. What is the percent yield for your efforts? a. 77.1% answer b. 22.9% c. 29.7% d. 130% e. 53.9% #14. (5 pts) Complete the reaction by balancing and indicating the phases (or states) on the product-side. may You need to make use of the solubility-table (on page 8). Na2CO3(aq) + CuCl2(aq) a. 2 NaCl(aq) + CuCO3(s) answer b. 2 NaCl(aq) + CuO(s) + CO2(g) c. 2 NaCl(s) + CuCO3(aq) d. 2 Na(s) + Cl2(g) + CuCO3(s) e. Na2Cu(s) + Cl2O2(g) + CO(g) #15. (5 pts) Which of the following is an example of a weak base? a. HCl b. CO2 CHEM 1311/TAMUCC/J. M. Southard, PhD %Yield Solution: %Yield Actual yield (g) 100 Theoretical yield (g) 4.82 g 100 6.25 g 77.1% + 4 c. H2SO4 d. NH3 answer (ammonia) e. All of these are weak bases #16. (5 pts) Which of the following substance(s) is/are insoluble in water? You may want to make use of the solubility-table (on page 8). a. NaBr b. PbBr2 c. AgCl d. NH4NO3 e. Both b and c answer #17. (5 pts) Complete the reaction by balancing and indicating the phases (or states) on the product-side. Na2CO3(aq) + 2 HCl(aq) a. 2 NaCl(aq) + H2CO3(aq) b. 2 NaCl(aq) + H2O(l) + CO2(g) best answer c. 2 NaCl(s) + H2CO3(aq) I would accept this answer as well (but H2CO3 decomposes) d. 2 Na(s) + Cl2(g) + H2O(l) + CO2(g) e. NaHCO3(aq) + NaCl(aq) + HCl(aq) #18. (5 pts) What volume, in milliliters, of 0.125 M NaOH would be needed to neutralize 25.0 mL of 0.234 M HCl? Equation: CaVa = CbVb. a. 13.4 mL b. 3.13 mL c. 5.85 mL d. 46.8 mLanswer e. 25.0 mL CaVa = CbVb Ca = 0.234 M Va = 25.0 mL Cb = 0.125 M Vb = ? Vb + Solution: Ca Va Cb Vb (0.234 M ) (25.0 mL) 0.125 M 46.8 mL #19. (5 pts) What is the molarity of a solution made by dissolving 57.6 g of sodium iodide (NaI, FW = 149.90) in sufficient H2O to produce a total volume of 1.00 L? a. 8.63 M b. 0.384 M answer Solution: Molarity no. moles of solute Liter of solution M CHEM 1311/TAMUCC/J. M. Southard, PhD 5 c. 2.60 M d. 8.63 103 M e. 0.116 M 57.6 g NaI 1 mol NaI 149.90 g NaI 1.00 L sol' n 0.384 mol NaI L sol' n 0.384 M NaI #20. (5 pts) Determine the oxidation number of carbon in H2CO3. You may want to may use of the oxidation number table on page 8. a. C+ b. C2+ c. C3+ d. C4+ answer e. C6+ Solution: H = +1; O = -2; C = x 2(+1) + x + 3(-2) = 0 2+x6=0 x4=0 x = 4 (i.e. C4+) Part II. Problem Solving (50 Points Total) #21. (10 pts) What is the maximum mass of Ni(OH) 2 that could be prepared by mixing two solutions that contain 25.9 g NiCl 2 and 10.0 g NaOH, respectively? This is a limiting reagent type problem. For full credit you must show your work and express your answer with the appropriate number of significant digits. NiCl2(aq) + 2 NaOH (aq) Ni(OH)2(s) + 2 NaCl(aq) FW 129.60 Amount on hand: 1 mol NiCl 2 0.200 mol NiCl 2 129.60 g NiCl 2 1 mol NaOH 10.0 g NaOH 0.250 mol NaOH 40.00 g NaOH 25.9 g NiCl 2 FW 40.00 FW 92.71 FW 58.44 Amount needed: 0.200 mol NiCl 2 2 mol NaOH 1 mol NiCl 2 0.400 mol NaOH 0.250 mol NaOH 1 mol NiCl 2 2 mol NaOH 0.125 mol NiCl 2 (a) You have 0.200 mol NiCl2 and you need 0.125 mol NiCl2. Therefore NiCl 2 is in excess! (b) You have 0.250 mol NaOH and you need 0.400 mol NaOH. Therefore NaOH is limiting! Base maximum mass calculation (i.e. "theoretical yield") upon moles of the limiting reactant: 0.250 mol NaOH 1 mol Ni(OH) 2 2 mol NaOH 92.7 g Ni(OH) 2 1 mol Ni(OH) 2 11.6 g Ni(OH) 2 Answer: 11.6 g Ni(OH)2 #22. (10 pts) Provide (a) the total ionic equation and (b) the net ionic equation for the following balanced precipitation reaction. CdCl2(aq) + 2 NaOH(aq) Cd(OH)2(s) + 2 NaCl(aq) Total ionic equation: CHEM 1311/TAMUCC/J. M. Southard, PhD 6 Cd2+(aq) + 2 Cl-(aq) + 2 Na+(aq) + 2 OH-(aq) Cd(OH)2(s) + 2 Na+(aq) + 2 Cl-(aq) Net ionic equation: Cd2+(aq) + 2 OH-(aq) Cd(OH)2(s) #23. (10 pts) Sulfur dioxide, a pollutant produced by burning coal and oil in power plants, can be removed by reaction with calcium carbonate: 2 SO2(g) + 2 CaCO3(s) + FW 64.06 FW 100.09 O2(g) FW 32.00 2 CaSO4(s) + 2 CO2(g) FW 136.14 FW 44.01 (a) What mass of CaCO3 is required to remove 155 g of SO2? (b) What mass of CaSO4 is formed when 155 g of SO2 is consumed completely? Solutions: (a) 155 g SO2 1 mol SO2 64.06 g SO2 2 mol CaCO 3 100.09 g CaCO 3 2 mol SO2 1 mol CaCO 3 242 g CaCO 3 Answer: 242 g CaCO3 required (b) 155 g SO2 1 mol SO 2 64.06 g SO 2 2 mol CaSO 4 136.14 g CaSO 4 2 mol SO 2 1 mol CaSO 4 329.41 g CaSO 4 Answer: 329 g CaSO4 formed #24. (2 5 pts) (a) What is the pH of a 0.0145 M HCl solution? (b) What is the H+ concentration of a solution that has a pH = 1.36? Equations: pH = -log[H+] & [H+] = 10-pH. Solution (need log and antilog key on calculator): (a) pH = log(0.0145) 1.84 Answer: pH = 1.84 (b) [H+] = 10-1.36 0.0437 M Answer: [H+] = 0.0437 M #25. (10 pts) If 38.55 mL of HCl is required to titrate 2.150 g Na 2CO3 according to the following equation, what is the molarity of the HCl solution? Na2CO3(aq) + 2 HCl(aq) 2 NaCl(aq) + CO2(g) + H2O(l) FW 106.00 Solution: Determine the moles of HCl required then divide by the volume given. CHEM 1311/TAMUCC/J. M. Southard, PhD 7 2.150 g Na 2 CO 3 1 mol Na 2 CO 3 106.00 g Na 2 CO 3 0.03855 L sol' n 2 mol HCl 1 mol Na 2 CO 3 1.052 mol HCl L sol' n 1.052 M HCl Answer: 1.052 M HCl CHEM 1311/TAMUCC/J. M. Southard, PhD 8 ... View Full Document

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