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In an experiment, you combine 83.77 g of iron with an excess of sulfur and then heat the mixture to obtain iron(III) sulfide. 2Fe(s) + 3S(s) → Fe2S3(s) What is the theoretical yield, in grams, of iron(III) sulfide?

Respuesta :

Catya
Start by looking up the molar mass of each thing in the reaction.

Fe = 55.845  g/mol
S = 32.065 g/mol
Fe2S3 = 207.9 g/mol

Next, convert Fe mas to moles.

83.77 g /  55.845  g/mol = 1.5 mol Fe

Us the stoichiometry (mole ratio) to get to moles product.

2 Fe reactants makes 1 Fe2S3 product

1.5 mol Fe * (1 Fe2S3 / 2 Fe) = 0.75 mol Fe2S3

Now multiply buy the Fe2S3 molar mass to get its weight in grams.

0.75 mol Fe2S3 * 207.9 g/mol = 155.925 g

This is the amount you should theoretically obtain using that much iron.









Oseni

The theoretical yield of iron (II) sulfide would be 155.92 g

Theoretical yield

It is the total stoichiometric product from a reaction.

From the equation of the reaction:

2Fe(s) + 3S(s) → Fe2S3(s)

The mole ratio of Fe to Fe2S3 is 2:1

Mole of 83.77 g Fe = 83.77/55.85

                               = 1.4999 moles

Equivalent mole of Fe2S3 = 1.4999/2

                                            = 0.75 moles

Mass of 0.75 mole Fe2S3 = 0.75 x 207.9

                                             = 155.92 g

More on theoretical yield can be found here: https://brainly.com/question/2506978

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