Question

Report Sheet for Experiment 18: mini “Synthesis” Of An Antacid Congratulations! You have just been hired as a researcher by the PRUB Chemical Company. Because of the large number of people eating spicy foods (and often eating them too fast), this company has asked you to develop a new antacid. They have given you some information, but the rest you must discover for yourself. Your boss has suggested that aluminum hydroxide may be effective as an antacid. This compound can be produced by combining aluminum oxide and water. In order to do this reaction, you must first write out a balanced equation: Al Out H. O N(OH)3 (1) ALO3 +34.0 -2Al(OH)3 Wale BE In order to test this antacid properly, you need to make 6500 g of aluminum hydroxide. As you enter your lab to synthesize it, you realize that you need to make a few calculations first. How many grams of aluminum oxide will you need to start with in order to make 6500 g of aluminum 10. hydroxide? (Please show all work!) Grom Aloth x 1 mol Allots Imad Al₂O3, 102gfy 9 78, A16H) Zond ALCOH Trd A1,03 (2) =_4250 g aluminum oxide 8.83.83 10 41.66x1039 4250 How many grams of water will you need to start with in order to make 6500 g of aluminum hydroxide? (Please show all work!) 6500g aluminum hydroxide - 4250 a oxide (3) =_2250 g water 2250 9 How many molecules of water would this be equal to? (Please show all work!) 1750-182125 H=la (x2) H₂O= log 2250g. (4) = _ 7.53X1025 125/6023 YI0 = 752.88 x 1023 Now that you have made these calculations, you are ready to do your reaction and make your antacid. To your horror, as you enter the lab you discover that the bottle of aluminum oxide is 1,53 completely empty. After a few moments of frustration, you realize that you can make your own aluminum oxide by combining oxygen gas with aluminum metal. Write out a balanced equation for this reaction: molecules of water O-Von (1)

Answer 1

1) Al2O₃ + 3 H₂O - QAL.COM) . M.W= 101.96gmał m. W = 78 gmal" mars of Allona = 6500g Al(OH) = 6500g - = 83.33 mol gmoli = 83. - Imol Al2O3 needed To make a mol Alcon imol Allons synthesis I mol Al2O₃ a needs 83.33 mol 83.33 mol dl2O3 - Allons mill need Moles of oly Oy = 33.33 41.665 mol Mars of slaOz = Mole x molar man = 41.665 mofe x 101.96gmoto = 4248.1634 g Mars of Al O₂ needed a 4248.29

- 1 To prepare 6500g alcoms , water needed = ? Moles of Allons = 83.33 mol To prepare a mol of ALLOH)3 - 3 moles of H₂O I are required for imol Alloh) - 3 mol H₂O will be required 83.33 mol Alloh, will require - 83.33 x 3 mol H₂O x Moles of H₂O= 83.33 x 3 mol = 124.995 mol - 125 mol Mars of H₂O = Molex Molar mass = 125 mol x 18 gmely = 2250 g Mars of H2O = 2250 g Molecules of M₂0 = Mole x Avogadro number = 15x6.022 x 1023 ::= 752-75x1023 = 7.53x1025 molecules

Balanced equation for synthesis of aluminuim ouide 4 Al +30 — 242 03