NAME: Exam 3 Introduction to General Chemistry Chem23/25 Spring 2011 Please write your name on every page. This exam is a closed-book, closed-notes exam. No outside material may be used and you may not discuss the exam with anyone else. The exam has 10 questions for a total of 82 points. The exam duration is 21/2 hours. Good luck! Useful relationships: 1.0 atm = 760 mm Hg 0 °C = 273 K For grading purposes only: Pages Problems Points Grader 1 1 - 2 ____ ____ 2 3 - 5 ____ ____ 3 6 - 7 ____ ____ 4 8 - 9 ____ ____ 5 10 ____ ____ TOTAL ____ points _____ %NAME: 1 1. Solid copper(II) oxide reacts with dilute hydrochloric acid to give a solution of copper(II) chloride and water. (a) Write a balanced equation for this reaction. Include state symbols. (4 points) CuO(s) + 2 HCl(aq)  CuCl2(aq) + H2O(l) (b) 7.95 g of copper(II) oxide are reacted with 250 mL of 1.0 M hydrochloric acid. Which is the limiting reagent? (4 points) 7.95 g CuO = 0.1 mol 250 mL of 1.0 M HCl = 0.25 mol therefore CuO is the limiting reagent (c) What is the theoretical yield (in grams) of copper(II) chloride from the reaction described in (b)? (4 points) 7.95 g CuO = 0.1 mol 0.1 mol CuO gives 0.1 mol CuCl2 0.1 mol CuCl2 weighs 13.5 g 2. When carbon dioxide is bubbled through a solution of lithium hydroxide, lithium carbonate is formed: 2 LiOH(aq) + CO2(g) → Li2CO3(aq) + H2O(l) (a) Balance the equation. (4 points) (b) How much lithium carbonate can be produced from 2.3 g of lithium hydroxide if the reaction goes in 95 % yield? (4 points) 2.3 g LiOH = 0.096 mol for 100% yield 0.096 mol LiOH gives 0.048 mol Li2CO3 for 95% yield 0.096 mol LiOH gives 0.046 mol Li2CO3 0.046 mol Li2CO3 = 3.37 gNAME: 2 3. One mole of an ideal gas at STP (0 °C and 1.0 atm pressure) occupies 22.4 L. What is the value of the gas constant, R in units of L.atm/mol.K ? (4 points) PV = nRT R = PV/nT R = (1.0)(22.4)/(1.0)(273) = 0.0821 L.atm/mol.K 4. When solid sodium carbonate is added to dilute hydrochloric acid the products are water, carbon dioxide, and aqueous sodium chloride. (a) Write a balanced equation for this reaction. Include state symbols. (4 points) Na2CO3(s) + 2 HCl(aq)  2 NaCl(aq) + H2O(l) + CO2(g) (b) 10.6 g of sodium carbonate are reacted with excess hydrochloric acid at 20 °C and 1 atm pressure. What volume of carbon dioxide is produced? (see Q3 for useful information) (6 points) 10.6 g Na2CO3 = 0.10 mol 0.10 mol Na2CO3 gives 0.10 mol CO2 V = nRT/P V = (0.10)(0.0821)(273+20)/(1.0) = 2.41 L 5. When solid ammonium nitrate is heated, nitrous oxide gas N2O is produced. In an experiment, N2O was collected over water. The total volume of gas produced was 500 mL at a pressure of 760 mm Hg and a temperature of 20.0 °C. How many moles of N2O were produced? The vapor pressure of water is 17.5 mm Hg at 20.0 °C and the gas constant R = 62.4 L.mmHg/mol.K (6 points) P = p(N2O) + p(H2O) P = 760 mmHg p(H2O) = 17.5 mmHg p(N2O) = 742.5 mmHg n = PV/RT n = (742.5)(0.5)/(62.4)(273+20.0) = 0.203 mol N2ONAME: 3 6. Aluminum metal reacts with excess sodium hydroxide to produce hydrogen as follows: 3 Al(s) + 6 NaOH(aq) → 2 Na3AlO3(aq) + 3 H2(g) (a) Balance the equation. (4 points) (b) How many moles of hydrogen gas are produced from 2.7 g aluminum (assume 100% yield)? (4 points) 2.7 g Al = 0.1 mol 0.1 mol Al gives 0.1 mol H2 (c) What volume of hydrogen gas is produced from 2.7 g of aluminum at 1 atm pressure and 20 °C? (see Q3 for useful information) (3 points) V = nRT/P V = (0.1)(0.0821)(273+20)/(1.0) = 2.41 L (d) What is the volume of the hydrogen gas in (c) if it is compressed to 200 atm at 40 °C? (3 points) P1V1/T1 = P2V2/T2 V2 = P1V1T2/P2T1 V2 = (1.0)(2.41)(273+40)/(200)(273+20) = 0.0128 L 7. Suggest the dominant intermolecular forces in each of the following molecules. (2 points each, 6 points) (a) (b) (c) ethane methanol fluoromethane London hydrogen bonding dipole-dipole H CHHCHHHH CHHO HH CHHFNAME: 4 8. Borane (BH3) and ammonia (NH3) have roughly the same molecular weight. Use Lewis structures and VSEPR theory to predict which has the higher boiling point. (8 points) Borane is trigonal planar, and so does not have a permanent dipole moment. Ammonia has a tetrahedral electronic geometry, and has a permanent dipole moment. Intermolecular forces will be stronger in ammonia, hence more energy will be required to move a molecule from the liquid to the vapor phase. Consequently ammonia has the higher boiling point. 9. The thermite reaction is used to produce molten iron for welding railroad tracks. Fe2O3(s) + 2 Al(s) → 2 Fe(s) + Al2O3(s) ΔH = -851 kJ How much energy is produced when 100.0 g Fe(s) is formed? (4 points) 100.0 g Fe = 1.79 mol Formation of 2.0 mol Fe gives out 851 kJ Formation of 1.79 mol Fe gives out 760 kJ HBHHNHHHNAME: 5 10. (a) Sketch a heating curve (energy on the x-axis, temperature on the y-axis) for mercury as it goes from solid to gas. Show the regions where the solid or liquid increases in temperature as energy increases, and show the phase changes. (4 points) Tb liquid Tm solid melting evaporating heating heating (b) How much energy is needed to vaporize 5.0 g of mercury? Assume a starting temperature of 20 °C. Useful relationships are q = mΔH and q = mcΔT. (6 points) Specific heat of Hg(s) c = 0.141 J/g°C Specific heat of Hg(l) c = 1.25 J/g°C Specific heat of Hg(g) c = 1.04 J/g°C Enthalpy of fusion of Hg ΔHfus = 11.6 J/g Enthalpy of vaporization of Hg ΔHvap = 29.5 J/g Melting point of Hg -39.0 °C Boiling point of Hg 357.0 °C (i) liquid at 20 °C to liquid at 357 °C q1 = (5.0)(1.25)(357-20) = 2106.25 J (ii) liquid at 357 °C to vapor at 357 °C q2 = (5.0)(29.5) = 147.50 J total energy = q1 + q2 = 2106.25 + 147.50 = 2254