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)

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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metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and   conduct electricity easily   while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

if it's not useful, feel free to report for mistake in answer.