Question

Determine the percent yield of a reaction that produces 28.65 g of Fe when 50.00 g of Fe2O3 react with excess Al according to the following reaction. Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s)

Answer 1

The answer to your question is 81.86 %

Step-by-step explanation:

Data

Percent yield = ?

mass of Fe = 28.65 g

Fe₂O₃ = 50 g

mass of Al = excess

Balanced chemical reaction

Fe₂O₃(s) + 2 Al (s) ⇒ Al₂O₃ (s) + 2 Fe (s)

Process

1.- Calculate the molar mass of the reactants

Fe₂O₃ = (56 x 2) + ( 16 x 3 ) = 112 + 48 = 160 g

Al = 2 x 27 = 54 g

2.- Calculate the theoretical yield

160 g of Fe₂O₃(s) --------------- 112 g of Fe

50 g of Fe₂O₃(s) --------------- x

x = (50 x 112) / 160

x = 35 g

3.- Calculate the percent yield

Percent yield = 28.65/35 x 100

Percent yield = 81.86 %