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

This preview shows page 4 - 5 out of 5 pages.

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
Image of page 4
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
Image of page 5

You've reached the end of your free preview.

Want to read all 5 pages?

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern

Stuck? We have tutors online 24/7 who can help you get unstuck.
A+ icon
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes