Hydrogen fluoride is used in the manufacture of Freons which destroy ozone in

Hydrogen fluoride is used in the manufacture of

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Hydrogen fluoride is used in the manufacture of Freons (which destroy ozone in the stratosphere) and in the production of aluminum metal. It is prepared by the reaction CaF 2 + H 2 SO 4 CaSO 4 + 2HF In one process 6.00 kg of CaF 2 are treated with an excess of H 2 SO 4 and yield 2.86 kg of HF. Calculate the percent yield of HF. (5 points) a. CaF2 + H2SO4 ----> CaSO4 + 2 HF... is balanced b. moles CaF2 = 6.00kg x (1000g / 1kg) x (1 mole / 78.07g) = 76.85 moles c. CaF2 is the limiting reagent.. that was given in the problem statement. d. from the balanced equation, 1 mole CaF2 ---> 2 moles HF... therefore... e. 76.85 moles CaF2 x (2 moles HF / 1 mole CaF2) = 153.7 moles HF f. 153.7 moles HF x (20.01g HF / 1 mole HF) = 3075.54 HF = 3.075kg HF g. % yield = 2.86 / 3.075kg x 100% = 93% (Reference: Chang 3.89) 10. Certain race cars use methanol (CH 3 OH; also called wood alcohol) as a fuel. Methanol has a molecular mass of 32.0 g/mol and a density of 0.79 g/mL. The combustion of methanol occurs according to the following equation: 2CH 3 OH + 3O 2 2CO 2 + 4H 2 O In a particular reaction 2.00 L of methanol are reacted with 80.0 kg of oxygen. a. What is the limiting reactant? (5 points) b. What reactant and how many grams of it are left over? (5 points) c. How many grams of carbon dioxide is produced? (5 points) i. Density = mass/volume ii. Mass of Methanol in the 2.00 liter of Methanol = 0.79 x 2000 = 1580g iii. 32.0g of methanol= 1 mol iv. 1580g of Methanol= 49.37 mol v. 2 CH3OH + 3 O2 = 2 CO2 + 4 H2O vi. 2 mol of Methanol reacts with 3 mol of oxygen vii. 49.37 mol of Methanol reacts with 3 x 49.37/2 = 74.055 mol of Oxygen 6 Copyright © 2017 by Thomas Edison State University . All rights reserved.
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viii. 1 mol oxygen = 32g ix. 74.055 mol oxygen = 32 x 74.055 = 2369.7g oxygen x. 80 kg = 80000g of oxygen is available xi. 80000 - 2369.76 = 77630.24g of oxygen will left unused xii. 2 mol of Methanol produces 2 mol of CO2 xiii. 49.37 mol of Methanol will produce 49.37 mol of CO2 xiv. 1 mol CO2= 44g of CO2 xv. 49.37 mol of CO2= 2172.28g of CO2 11. An iron bar weighed 664 g. After the bar had been standing in moist air for a month, exactly one-eighth of the iron turned to rust (Fe 2 O 3 ). Calculate the final mass of the iron bar and rust. (5 points) a. Fe + O2 → Fe2O3 b. 4Fe + 3O2 → 2Fe2O3 c. 1 mol Fe2O3 = 159.687g d. 1 mol Fe = 55.845g e. 664g/8 = 83g; 664g-83g = 581g Fe f. 83g Fe x (1 mol Fe/ 55.845g) x (1 mol Fe2O3 / 2 mol Fe) x (159.687g Fe2O3 / 1 mol g. Fe2O3 = 118.668 g Fe2O3 (Reference: Chang 3.111) 7 Copyright © 2017 by Thomas Edison State University . All rights reserved.
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  • Spring '14
  • Written Assignment, Sulfuric acid, Thomas Edison State University

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