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City College, Chemistry Department Chemistry 10301, sections H*, Prof. T. Lazaridis Second Midterm exam, Nov 5th, 2014 Last Name: ____________________________________________ First Name: _____________________________________________ Note: There are 10 questions in this exam (check both sides of the sheet). Fill in your answer in the blank space provided immediately following each question. 1/2 point will be subtracted every time you report a numerical result with an incorrect number of significant figures. The last sheet of this exam contains data that may be needed to answer these questions. R= 0.08206 atm L/mol K. 1. (10) Using data in your datasheet, predict whether a reaction will take place by mixing the following substances. If so, give the type of reaction and write a net ionic equation for the reaction. a. (5) BaS (aq) and CuSO4 (aq) Precipitation. Ba2+ (aq) + S2- (aq) + Cu2+ (aq) + SO42- (aq) ----> BaSO4 (s) + CuS (s) b. (5) Cr(OH)3 (s) and HBr (aq) Acid-base neutralization Cr(OH)3 (s) + 3H+ (aq) ------> Cr3+ (aq) + 3 H2O (l) 2. (8) a) (5) Give the oxidation numbers of each element in the reactants and products of the reaction CH4 + 2 NO2 ------> N2 + CO2 + 2 H2O . b. (3) Identify the oxidizing and the reducing agent in this reaction. a. Reactants Products C -4 +4 N +4 0 H +1 +1 O -2 -2 b. Oxidizing agent is NO2, reducing agent is CH4. 3. (12) A sample of NaCl weighing 1.4477 g is dissolved in water and diluted to 250.0 mL. What volume of this solution will be needed to titrate 25.00 mL of 0.1000 M AgNO3 (aq)? NaCl (aq) + AgNO3 (aq) ----> AgCl (s) + NaNO3 (aq) Moles of AgNO3: 25.00 X10-3 L X 0.1000 M = 25.00 X 10-4 We need the same number of moles of NaCl. We have 1.4477 g/58.442 (g/mol) = 0.024772 moles in 250.0 mL 25.00 X 10-4 moles X 250.0 mL/0.024772 mol = 25.23 mL 4. (10) A gaseous hydrocarbon has a density of 1.69 g/L at 24 oC and 743 torr. Combustion analysis revealed that its empirical formula is CH2. Find the molecular formula of this hydrocarbon. From the ideal gas law we find the molar mass: 1.69 g/L * 0.08206 L atm/mol K * (24+273) K M=dRT/P = ---------------------------------------------------------- = 42.13 g/mol 743 torr X 1 atm/760 torr The empirical formula mass is 14.026. 42.13/14.026= 3 So, molecular formula is C3H6. 5. (10) How many liters of CO2 (g) can be produced in the reaction of 5.24 L CO (g) and 2.65 L O2 (g) if all three gases are measured at the same temperature and pressure? 2 CO (g) + O2 (g) ------> 2 CO2 (g) If all gases are measured at the same T,P, ratio of moles=ratio of volumes. Find the limiting reactant: 5.24 L CO X (2molCO2/2molCO) = 5.24 L CO2 ß less 2.65 L O2 X (2molCO2/molO2) = 5.30 L CO2 So, CO is limiting. The actual amount produced will be 5.24 L. 6. (15) A 2.02 g sample of Al reacts with excess HCl (aq) and the liberated hydrogen is collected over water at a temperature of 24 oC. What is the total volume of the gas collected at a barometric pressure of 760.0 torr? (The vapor pressure of water at 24oC is 22.4 torr). The reaction is: 2 Al (s) + 6 H+ (aq) -> 2Al3+ (aq) + 3 H2 (g) First find the moles of H2 produced: 2.02 g Al/26.98 (g/mol) X (3/2) = 0.112 mol The total pressure is 760.0 torr. The partial pressure of H2 is 760.0-22.4 torr. The volume is found from the IG law: V=nRT/P = 0.112 X 0.08206X(273+24) / (760-22.4) torr X 1 atm/760 torr= = 2.82 L 7. (15) Calculate the standard enthalpy of the reaction BrCl(g) --------> Br (g) + Cl (g) using the following data: Br2(l) -----> Br2 (g) ΔHo = + 30.91 kJ Br2(g) ------> 2Br (g) ΔHo = +192.9 kJ Cl2(g) ------> 2Cl (g) ΔHo = +243.4 kJ Br2(l) + Cl2(g) -----> 2 BrCl(g) ΔHo = + 29.2 kJ Invert the last equation and multiply by ½. Multiply all others by ½ and add. ΔHo= -½*29.2 + ½*30.91 + ½*192.9 + ½* 243.4 = 219.0 kJ 8. (5) A 0.828-g sample of gasoline is burned in a bomb calorimeter with a heat capacity of 9.89 kJ/K. The temperature in the calorimeter rises from 22.75 to 26.37 oC . Calculate the heat of combustion (ΔHo) of the gasoline, in kJ per gram. q= hc X ΔT = 9.89 kJ/K X (26.37-22.75) K = 35.8 kJ heat of combustion: -35.8 kJ/0.828 g = -43.2 kJ/g 9. (5) Use data from your datasheet to determine the standard enthalpy change at 25 oC for the reaction NH4+ (aq) + OH- (aq) ------> NH3 (g) + H2O (l) ΔHo= {enthalpy of formation of products}-{enthalpy of formation of reactants}= ΔHof (NH3) + ΔHof (H2O(l)) – ΔHof (NH4+) – ΔHof(OH-) = -46.11 +(-285.8) - (-132.5) - (-230.0) = 30.6 kJ 10. (10) How many liters of ethane, measured at 17oC and 714 torr, must be burned to give off 2.75X104 kJ of heat? 2 C2H6 (g) + 7 O2 (g) ------> 4 CO2 (g) + 6 H2O (l) ΔH=-3.12X103 kJ Find moles of ethane: 2.75X104 kJ X 2 mol/ 3.12X103 kJ = 17.6 mol Use IG law to find the volume: V=nRT/P = 17.6 mol*0.08206 atmL/molK*(17+273) /714 torrX 1atm/760torr=446 L