Apply In an experiment, you combine 83.77 g of iron with an...

Apply In an experiment, you combine 83.77 g of iron with an excess of sulfur and then heat the mixture to obtain iron(lill) 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?

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00:02 Hi there.

00:03 In this question, we are trying to determine the theoretical yield in grams of the product here, the iron 3 sulfide.

00:11 So we are trying to find grams of this product.

00:15 We know we are starting with 83 .77 grams of iron and an excess of sulfur.

00:24 All right.

00:25 So this is a mass -to -mass stochometry problem.

00:29 We already have the balanced equation.

00:30 It was given to us.

00:32 So we need to convert from grams of the reactant iron to grams of the product.

00:37 So to do that, we need to go through the mole because we have to use the mole ratio.

00:43 So our first step is going to be to convert from grams of iron to moles of iron.

00:50 We're going to do that by using its molar mass.

00:53 Then we can convert from moles of iron to moles of the fv2s3 by using the mole ratio.

01:02 And the mole ratio comes from the coefficients in that balanced equation.

01:07 So we see that there are 2fe for every 1 f .e .2 .s .3 that's produced.

01:14 Since there's nothing written in front of f .e .2 .s .3, it's understood that it's a 1 for the coefficient.

01:19 So we have a 2 to 1 mole ratio there.

01:23 And then we can convert back to grams of fe2s3 by using its molar mass.

01:29 So it looks like we can solve this in three steps...

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