Question

Apply In an experiment, you combine 83.77 $\mathrm{g}$ of iron with an excess of sulfur and then heat the mixture to obtain iron(lil) sulfide. $$2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{S}(\mathrm{s}) \rightarrow \mathrm{Fe}_{2} \mathrm{S}_{3}(\mathrm{s})$$ What is the theoretical yield, in grams, of iron(III) sulfide?

          Apply In an experiment, you combine 83.77 $\mathrm{g}$ of iron with an excess of sulfur
and then heat the mixture to obtain iron(lil) sulfide.
$$2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{S}(\mathrm{s}) \rightarrow \mathrm{Fe}_{2} \mathrm{S}_{3}(\mathrm{s})$$
What is the theoretical yield, in grams, of iron(III) sulfide?

Added by Mar-A W.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Madhur L

05/18/2023

Step 1

First, we need to find the moles of iron (Fe) in 83.77 g. The molar mass of iron is 55.85 g/mol. $$\text{moles of Fe} = \frac{83.77 \text{ g}}{55.85 \text{ g/mol}}$$ Show more…

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Apply In an experiment, you combine 83.77 $\mathrm{g}$ of iron with an excess of sulfur and then heat the mixture to obtain iron(lil) sulfide. $$2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{S}(\mathrm{s}) \rightarrow \mathrm{Fe}_{2} \mathrm{S}_{3}(\mathrm{s})$$ What is the theoretical yield, in grams, of iron(III) sulfide?