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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 :

joseaaronlara

Answer:

The answer to your question is 81.86 %

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 %

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