Question

please answer questions 1-8, I don't know what i'm doing wrong :(

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: aluminum hydroxide (1) Al₂ O₃ + 3H₂O - 2A1(OH)₂ aluminum water oxide 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 hydroxide? (Please show all work!) Al2O3 + 3H2O → 2 Al(OH)2 1 mol ALO3 x 101.96 Al2O3 2 mol ALLOH) 1. mol hizo 4500g Allots) x 2 mol ALCOB 77.98g Allohe 132548D 155 96 = 849 8.85 g aluminum oxide

Answer 1

- 83.3mol. © Allzos +3H0 - 2A (OH)2 apangs of AllOH) = 6500g. mole " 2 6500g . molastmass of A16H), in g/mol z 65oog 97-98 g/mol so 2 mol of AlloH) Requine 1 mol of Alzos. so 03:36 " " " " *83.35 mol of Alz oo spame of Al2O3 = ( x 83.35 mol x molar mass of Alloy *83.3.5mpx 101.96 g/mol - 4249.183 9. . & again . 3 mol of H2o required to produce a mol A16H), : so, to produce 83.35 mol of Alloh, Emol. A16H), — 3 mot of H2o- 83.35 1 1 (2*8234) mol of energy groanas of water = 3 x 83.35 mol & 18 5 mol so 9 molecules a 2250°45 g of water = given mase x 6.022 x 1023 molan massa = 2250-45v6.022 x 1023. 18. : s 2810?h 7.5 3 X 10. molecules of water. i

I MAI of 30 2Al2O3 1 6 so 2 mol of Al₂O3 is pequipad u mol of Al. lá (4-2) mot of Al 2 I so (A x83:35) mol of Ako nequired ax 1 x83.35 mol of Al - 83.35 mol 8 AM I son grams of Al metal = 83.35 molx 26.98g/mol 3 2248-7839 of Al imot of Al₂O3 required 3 - 2 mol of O2 **83.35."" " " (3.44*33:35) mot q ºz. so growns of o2 = 2 x 33 35 x 32 g/mol :: 2000.4g. ③ mots of oxygen = given mass - R - molan mass 220004 g .662-5125 mol 302 g/mory -