Question

For the following reaction, 55,7 grams of iron(III)oxide are allowed to react with 17.3 grams of aluminum iron II)oxides) aluminum's) - aluminum oxide(s) + iron(s) grams What is the maximum amount of aluminum oxide that can be formed? What is the FORMULA for the limiting reagent? grams What amount of the excess reagent remains after the reaction is complete? SA Retry Entire Group more group attempts remaining RT F G H

Answer 1

fez 03 (3) + Al(s) - Al₂O(S) + fero Molar mass of fe, O3= 231153 g/mol Polen mass of Al 21 g/mol. Molar mass of Al O3 = 101.96 gland molar mass of tee 55.8 g/mol Exz lots 2419 x AlB345) +fers de Oz (s) + 2 Alis) – Alo (s) +2 te Иus 231.59 2x27 2547 rozg zg 2x57 21122 231.57 of Fez reacts with 54g of Al 55.78 of Foz will react inth 54 x 55:7 - 12.9 g of Al As there is 17.39 of Al. Al is in exess. 231-5

54g of Al reacts with 231.5g of 17.3 g of Al will react with 231.5 X 17.3 174. 16g of feron As there is only 55 Tg of fezos. fesz is limiting reagent. * 4, 231.5 g of fez gives loag of Al 03 sul 55.7g of felos will give 102 x 55-7-24:57 of 41208 231.5 Exuss At reagent is Al. i is in excess by amount o o 17.3-12 19 = 4:48. Al is in excess by 4-4g