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?

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.