2 G Of Fe example essay topic

Gayle Adkinson Director of Operations Adkinson Chemical Corporation Dear Mrs. Adkinson: A sample of copper has been produced as requested. We are pleased to report that the most copper has been produced through the most efficient method. But we regrettably to report that an accident had occurred during the process (which you will not be billed for). This accident led to a series of investigations to weed out incompetent workers. The remaining technicians were the best and most capable of producing the highest purity of copper at the minimum cost. Captain Copper Chemical Corporation supplies application-specific process chemicals for most of the science industry.

The combination of high technology products and in-depth support, from the technical service team, allows the customer to maximize productivity and minimize cost. Captain Copper Chemical Corporation is a technology-driven, not marketing-driven company. We pride ourselves on our capacity to solve clients chemical process issues, and will develop custom products, when required. Captain Copper Corporation is currently both the leading North American distributor of chemicals and a world leader in manufacturing a variety of chemical specialties. In locations throughout the world, youll find Captain Copper Chemical Corporation operations upholding standards that keep safety, health and environmental considerations a top priority. OSHA has deemed the of using safe techniques and proper disposal of wastes. strives to be the most dependable supplier of high-quality goods and services for our important constituents. is dedicated into achieving mutually profitable relationships with our customers and suppliers by providing quality products and services, on time and consistently, and by remaining constant aware of changing customer needs.

Captain Chemical Corporation invites you to share your chemical process problems, and issues, so that we may offer and demonstrate our capacity to assist you in becoming the leader in your industry. Procedure Day 1 1. Determine limiting reactant 2. Prepare a data table with the following: mass of- beaker A, beaker B, beaker A + CuSO 4.5 H 2 O, beaker B + Fe, CuSO 4.5 H 2 O, Fe, filter paper, beaker B + filter paper + Cu, Cu, and observations 3. Clean and dry two 250 mL beakers.

Label one A and the other B. Mass each beaker. Record results in data table 4. Purchase 10 g of CuSO 4.5 H 2 O and 2.22 g of Fe (less than needed to make sure all Fe reacts). Place CuSO 4.5 H 2 O in beaker A and Fe in beaker B. Mass each beaker.

Record results in data table 5. Dissolve CuSO 4.5 H 2 O (Dr. Merck informed us that CuSO 4.5 H 2 O was soluble) 6. Heat beaker A (w / CuSO 4.5 H 2 O (aq) ). Heat until solution is hot but not boiling. 7. Remove beaker A from hot plate using beaker tongs and pour CuSO 4.5 H 2 O (aq) into beaker A and let CuSO 4.5 H 2 O react with Fe 8.

Let the chemicals react. Done for the day. Day 2 9. Make observations of solution and precipitate. 10. Setup to filter solution.

Set ring stand with funnel. Mass filter paper. Record results in data table. Fold filter paper and set it up in funnel. Wet filter paper to remove all space between filter paper and funnel.

11. Decant solution and filter into beaker A using stirring rod. Rinse precipitate (Cu) with distilled water. Filter solution again. Do this about three times. 12.

Dispose of filtrate (FeSO 4). Place filter paper into beaker B with precipitate. Dry beaker B. Done for the day. Day 3 13. Make observations of copper.

Scrape any remaining copper on filter paper into beaker. Mass the copper and record results into data table (Dr. Merck recommends removing all iron impurities with a magnet) Data Table The following calculations were made prior to the lab for limiting reactant determination to make efficient use of all materials. molar mass of CuSO 4.5 H 2 O = 63.5 g Cu + 32.1 g S + 144.0 g O + 10.1 g H = 249.7 g molar mass of Fe = 55.8 g molar mass of FeSO 4 = 55.8 g Fe + 32.1 g S + 64.0 g O = 151.9 g FeSO 4 molar mass of Cu = 63.5 g Cu mols of 10 g CuSO 4.5 H 2 O = 0.040048058 mols of CuSO 4.5 H 2 O mols of 10 g Fe = 0.17921147 mols of Fe needed if 10 g CuSO 4.5 H 2 O mols of CuSO 4.5 H 2 O if 10 g Fe mass of Fe needed if 10 g of CuSO 4.5 H 2 O is used: = 2.234681618 g Fe = 2.23 g Fe mass of CuSO 4.5 H 2 O needed if 10 g of Fe is used: = 44.74910394 g CuSO 4 = 44.7 g CuSO 4 mass of CuSO 4.5 H 2 O CuSO 4.5 H 2 O is the limiting reactant. mass of Cu expected to be produced from 2.22 g of Fe: = 2.526344086 g Cu = 2.53 g Cu The following calculation was made after the lab for percent yield determination: 2.50 g Cu (actual mass) 2.53 g Cu (theoretical mass) x 100 = 98.8142292% = 99% 99% of the theoretical mass of copper to be produced was retrieved. Materials CuSO 4.5 H 2 O. 5 grams = $. 5010 g = $10.00 Fe. 5 grams = $1.002. 22 g x 2 = $8.88 Total$18.88.