Question

For the following reaction, 37.0 grams of aluminum oxide are allowed to react with 133 grams of sulfuric acid . aluminum oxide(s) + sulfuric acid(aq) → aluminum sulfate(aq) + water(1) What is the maximum amount of aluminum sulfate that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams

Answer 1

SUL Sol tron) Ay@z +34,5 - Alq CSQ]3+ 340 cilen 379 Abg 1339 th so Now mores of neu - 379 0.362 mol 101.969/mol moies of M_38 = 1339 z 10 356 mol 98.079 | Alio is limiting Rogent 1 now from Earation Imole of Al2O3 - gives i mole Aluminum sulphate gives 0.362 moles of Thon 0.362 mol of 100 y Ab (50)3 Ah - Call 0.362 mol Maximum amount of Aluminum sufole that con beformed < = 0.362 mb1 x str 2169 and 123.859 mars= molal mass races GF A1604/3 of 11689,1) Now imole of Aho Realices - 3 mole of Hala Then 0.362 mol Aho ? -0.362 x 3 moes of H IG 10086 moles - 1.356 - 1.086 Remaining 0.27 moles a mass = 0.27 mol x 98.07g/mol of excess . = 26.47 glomy Reagent Reanies mow of its