4 2HF 781g 40g Theoretical yield of HF 600 x 40 781 307 31 kg HF yield of HF

# 4 2hf 781g 40g theoretical yield of hf 600 x 40 781

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4 + 2HF 78.1g 40g Theoretical yield of HF = (6.00 x 40) / 78.1 = 3.07 = 3.1 kg HF % yield of HF = (2.86 / 3.1) x 100% = 92% Molar mass of CaF 2 is 78.07 g/mol and molar mass of CaSO 4 is 136.14 g/mol Molar mass of HF is = 1.008 + 18.998 = 20.01 g/mol
1 mole of CaF 2 produces 2 mole of HF. So, 78.07 grams of CaF 2 produces 2 × 20.01 = 40.02 grams of HF Theoretical yield of HF = (6.00 × 40.02) / 78.07 = 3.07 kg of HF % yield of HF = (2.86 / 3.07) × 100 = 93.15 % 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) From the balanced equation 2 moles of CH 3 OH and 3 moles of O 2 required to produce 2 mole of CO 2 The molar mass of Methanol = 32.0 / 0.79 = 40.5 ml/mole Moles of Methanol CH 3 OH = (2000 × 1) / 40.5 = 40.9 moles Every 2 moles of CH 3 OH will react with 3 moles of O 2 Number of moles required of in reaction = (40.9 × 3 ) / 2 = 74.1 moles 74.1 moles of O 2 = (74.1 × 2 × 15.99) = 2369.71 grams of O 2 Since we have only 2.00 L methanol, make it as limiting reagent. b. What reactant and how many grams of it are left over? (5 points) Left over reactant will be Oxygen, Mass of left over oxygen = 80000 – 2369.71 = 77630.29 c. How many grams of carbon dioxide is produced? (5 points)’ Every 2 moles of CH 3 OH required to produce 2 mole of CO 2 Mass of CO 2 Produced = (40.9 × 44.0) = 2173.6 grams. 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) (Reference: Chang 3.111) The rusting process of iron bar can be given as follows: 4 Fe + 3O 2 2 Fe 2 O 3 Molecular mass of Fe in this reaction = 4 × 55.845 = 223.2 g/mol Molar mass of Fe 2 O 3 in this reaction = (4 × 55.845) + (6 × 15.99) = 319.2 g/mol Mass of Fe lost in conversion to rust = 664 / 8 = 83 grams Mass of rust (Fe 2 O 3) Produced in this reaction = (319.2 × 83) / 223.3 = 118.64 grams Final mass of iron bar (Fe) left = 664 – 83 = 581

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