Question

Determine the limiting reactant and percent yield of a reaction that produces 573 8 of Fe when 100.00 g of Fejreact with 50,00 g Alaccording to the following reaction Fe2O3(s) + 2Al(s) - Aly0315) + 2 Fe(s) 200* Paragraph BI VALT

Answer 1

Molar mass of fe2O3 = 159.69 g/mol Al2O3 + 2 fe Fe2O3 (1 mole) -759.69 OR OR + 2 Al (2 mole) 2826.489 = 53964 to balanced . According chemical equation 53.96 g So, I AL AL require require 159.69 g Fe2O3 159.69 g 53.96 159.69 X 500 g 53.96 A, 50.00g Al vrequire - +_147:97_9_Fe203. : 50 g Al require 147.97 g react completely, but we - 200 g fez O₂ . So f. 2.03 is Fe2O3 to have only limiting reagent.

so, amount of product depend upon the mass of limiting reagent. Again, fe2O3 +2Al Al 203 + 2 fe - (1 mole) - (2 mole) = 159-69 g = 2x55.8459 According to equation, 1. 159.69 g fe203 produce 111.69 g Fe so, 1 g Fe2O3 produce 111.69 g fe _ _159.69 8, 100 g fe203_produce 111.69._X100 g fe 159.69 = 111.69 g 7 69.94 g fe

Theoretically, amount offe, = 69.94 g. . The aretical Yeild 57:37 X100% 69.71 - = 81.9253_mon Percent Yeild = 81.92 %