Module 6 Unit 2 Assignment D

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Fanshawe College *

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MI-JUN ACE

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Chemistry

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Jan 9, 2024

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MODULE 6 UNIT 2 Assignment Part D new By: Connie Warburton Each answer – 4 marks. To get full 4 marks, show your work in each question. Include references. 1. If liquid carbon disulfide (CS 2 ) reacts with 450 mL of oxygen to produce the gases carbon dioxide and sulfur dioxide, what volume of each product is produced? CS 2 + 3O 2 CO 2 + 2SO 2 450 mL 0.450 L 0.450 L x 1 mol / 22.4 L = 0.0200892857 = 0.02009 mol O 2 0.02009 mol O 2 x 1 mol CO 2 / 3 mol O 2 = 0.0066966667 = 0.0067 mol CO 2 0.0067 mol CO 2 x 22.4 L / 1 mol = 0.15008 = 0.150 L CO 2 = 150 mL CO 2 0.02009 mol O 2 x 2 mol SO 2 / 3 mol O 2 = 0.0133933333 = 0.0134 mol SO 2 0.0134 mol SO 2 x 22.4 L / 1 mol = 0.30016 = 0.300 L SO 2 = 300 mL SO 2 2. Calculate the volume of F 2 gas at STP produced by the electrolysis of 8.2 g of KF. 2KF → 2K + F 2 1 mol KF = 39.098 + 18.998 = 58.096 g KF 8.2 g KF x 1 mol / 58.096 g = 0.14114456899 = 0.141 mol KF 0.141 mol KF x 1 mol F 2 / 2 mol KF = 0.0705 mol F 2 0.0705 mol F 2 x 22.4 L / 1 mol = 1.5792 =1.6 L F 2 3. Ammonium sulfate, an important fertilizer, can be prepared by the reaction of ammonia with sulfuric acid according to the following balanced equation: 2 NH 3 (g) + H 2 SO 4 (NH 4 ) 2 SO 4 (aq) 1
Calculate the volume of NH 3 (in liters) needed at 20ºC and 2533.12 kPa to react with 150 kg of H 2 SO 4 1 mol H 2 SO 4 = (2 x 1.008g) + 32.06g + (4 x 16.00g) = 98.076g 150 kg 150,000 g H 2 SO 4 (1.5 x 10 5 ) 150,000 g H 2 SO 4 x 1 mol / 98.076g = 1529.426159305 = 1.53 x 10 3 mol H 2 SO 4 1.53 x 10 3 mol H 2 SO 4 x 2 mol NH 3 / 1 mol H 2 SO 4 = 3.06 x 10 3 mol NH 3 3.06 x 10 3 mol NH 3 x 22.4 L / 1 mol = 68,544 L = 6.85 x 10 4 L NH 3 Vf = 6.85 x 10 4 L NH 3 x 293 K / 273 K x 101 kPa / 2533.12 kPa = 2,931.30598505 = 2.93 x 10 3 L NH 3 4. A 3.25 gram sample of solid calcium carbide (CaC 2 ) reacts with water to produce acetylene gas (C 2 H 2 ) and aqueous calcium hydroxide. If the acetylene was collected over water at 17ºC and 98.66 kPa, how many milliliters of acetylene were produced? CaC 2 + 2H 2 O C 2 H 2 + Ca(OH) 2 1 mol CaC 2 = 40.078g + (2 x 12.011g) = 64.1 g 3.25 g CaC 2 x 1 mol / 64.1 g = 0.0507020281 = 0.051 mol CaC 2 0.051 mol CaC 2 x 1 mol C 2 H 2 / 1 mol CaC 2 = 0.051 mol C 2 H 2 0.051 mol C 2 H 2 x 22.4 L / 1 mol = 1.1424 = 1.14 L C 2 H 2 Vf = 1.14 L C 2 H 2 x 290 K / 273 K x 101 kPa / 98.66 kPa = 1.2397110289 = 1.24 L C 2 H 2 1.24 L x 1000 mL / 1 L = 1240 mL C 2 H 2 5. If you burned one gallon of gas (C 8 H 18 )(approximately 4000 grams), how many liters of carbon dioxide would be produced at a temperature of 21.0°C and a pressure of 101.3 kPa? C 8 H 18 + 8O 2 8CO 2 + 9H 2 1 mol C 8 H 18 = (8 x 12.011g) + (18 x 1.008g) = 114.232g 4000 g C 8 H 18 x 1 mol / 114.232g = 35.0164577351 = 35.016 mol C 8 H 18 35.016 mol C 8 H 18 x 8 mol CO 2 / 1 mol C 8 H 18 = 280.128 mol CO 2 280.128 mol CO 2 x 22.4 L / 1 mol = 6274.8672 = 6275 L CO 2 Vf = 6275 L CO 2 x 294 K / 273 K x 101 kPa / 101.3 kPa =6736.6793985876 = 6738 L CO 2 6. A sample of magnesium with a mass of 1.00 g is burned in oxygen to produce an oxide with a mass of 1.66 g. What is the empirical formula of the magnesium oxide produced? 2
Mg 1.00 g / 24.305g/mol = 0.041144 mol O 1.66 g / 16.00g/mol = 0.04125 mol Mg/Mg 0.041144 mol / 0.041144 mol = 1 O/Mg 0.04125 mol / 0.041144 mol = 1 Empirical formula is MgO 7. How much heat will be released when 6.44 g of sulfur reacts with oxygen according to the following equation: 2S + 3O 2 2SO 3 + 791.4 kJ 1 mol SO 3 = 32.06 + (3 x 16.00g) = 80.06 g 6.44 g x 1 mol / 80.06g = 0.08043967 mol SO 3 0.08043967 mol SO 3 x 791.4 kJ / 1 mol = 63.65995484 = 63.7 kJ 8. What mass of propane, C3H8, must be burned in order to produce 76000 kJ energy? C 3 H 8 + 5O 2 3CO 2 + 4H 2 O + 2200 kJ 1 mol C 3 H 8 = (3 x 12.011g) + (8 x 1.008g) = 44.097g 76000 kJ x 1 mol / 2200 kJ x 44.097g / 1 mol = 1523.350909 = 1523.4 g C 3 H 8 9. How much heat will be absorbed when 13.7 g nitrogen reacts with oxygen according to the following equation? N 2 + O 2 + 180 kJ 2NO 1 mol N 2 = 2 x 14.007 = 28.014 g 13.7 g N 2 x 1 mol / 28.014 g = 0.489041193 = 0.489 mol N 2 0.489 mol N 2 x 180 kJ / 1 mol = 88.02 kJ 3
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References: Fanshawe College (2023) “ Module 6 Unit 2 Presentation” retrieved from https://olmoodle.ontariolearn.com/mod/scorm/player.php? a=8783¤torg=articulate_rise&scoid=17776&sesskey=XwW1XCqyvM &display=popup&mode=normal Ptable.com (2023) “periodic table” retrieved from https://ptable.com/image/periodic-table.svg 4