5.04a How many grams of H2S can be formed

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From the course by University of Kentucky

Chemistry

637 ratings

University of Kentucky

Chemistry

637 ratings

This course is designed to cover subjects in advanced high school chemistry courses, correlating to the standard topics as established by the American Chemical Society. This course is a precursor to the Advanced Chemistry Coursera course. Areas that are covered include atomic structure, periodic trends, compounds, reactions and stoichiometry, bonding, and thermochemistry.

From the lesson

Week 5

This unit will delve into the quantitative relationships we can determine from a balanced chemical equation to determine the relative amounts of substances needed to react or the amount of products formed. We will also explore limiting reagents and percent yield to address practical aspects of chemical reactions because there may be leftover reactants and/or incomplete formation of products.

5.01 Stoichiometry of Chemical Reactions: Mole to Mole 8:19

5.01a How many Moles of water are produced1:11

5.01b How many Moles of CO2 are produced 1:47

5.02 Stoichiometry of Chemical Reactions: Mass Relationships11:02

5.02a How many grams of magnesium are required 2:27

5.02b How many grams of magnesium do we need 4:34

5.03 Limiting and Excess Reagents11:18

5.04 Limiting and Excess Reagents: Calculations 7:57

5.04a How many grams of H2S can be formed 4:13

5.04b What mass of excess reactant will remain 6:31

5.05 Theoretical & Percent Yields 9:56

5.05a Methyl Salicylate is prepared5:48

5.05b A student needs 625 g of zinc sulfide5:32

5.06 Solution Concentration 15:16

5.06a Find the mass percentage for a solution prepared 1:56

5.06b Find the volume percentage for a solution prepared 1:51

5.06c Find the mass/volume percentage for a solution prepared 2:39

5.06d What is the molarity of a solution made 2:23

5.06e How many grams of LiCl do we need 2:44

5.07 Dilution of Solutions7:16

5.07a What is the concentration of the solution 1:45

5.08 Solution Stoichiometry 9:06

5.08a What volume of HBr is required 5:03

Meet the Instructors

Dr. Allison Soult
Lecturer
Chemistry
Dr. Kim Woodrum
Sr. Lecturer
Chemistry

If 10 grams of HCl is added to 11.4 grams iron sulfite,

how many grams of H2S can be formed?

I'm given a balanced chemical equation, which I'm going to need later on when I

do the calculation, and right now I'm going to put in all my molar masses for

each substance that way I have them there when I need them.

So FeS has a molar mass of 87.91 grams per mole,

HCl 36.46 grams per mole,

FeCl2 126.75 grams per mole,

and 34.08 grams per mole.

So I may not actually need all of these,

but given that several substances are named,

I thought it would be easier just to start with the molar masses of everything so

then you don't have to stop mid-problem solving and go back and find that value.

So I'm also going to write down the information that's given in the problem.

So we're told we have 10 grams of HCl and

11.4 grams of FeS and how many grams of H2S?

I'm going to put a question mark there to remind me

that that's what I'm looking for.

Now, notice that in the problem I'm given the amounts of two different reactants.

And that's my big red flag that this is a limiting reagent problem.

So I'm actually going to have to solve two problems in order to

figure out the actual answer to the problem.

So the first thing I'm going to do is I'm going to assume

that FeS is my limiting reagent.

And I'm going to do the calculation assuming that it's limiting.

So I'm going to start with the 11.4 grams of FeS.

I'm going to use my molar mass, 87.91 grams per mole.

I'm going to look at my mole to mole ratio between the FeS and the H2S because

remember that's what I'm looking for is that mass of the H2S for the product.

So I see that for every 1 mole of FeS, I'm going to produce 1 mole of H2S.

And then I'm going to go ahead and convert that to grams,

34.08 grams per mole of H2S.

And I put the mole on the bottom because I want my units to cancel out.

So grams of FeS cancels.

Moles of FeS.

Moles of H2S and what I'm left with are grams of H2S.

And again, this is assuming that the iron sulfide is the limiting reagent.

When I do the calculation, I get 4.42 grams of H2S.

Now remember, I don't know if this is my answer because I don't know which

species is my limiting reagent.

I've assumed iron sulfide is limiting, now I need to do a similar calculation this

time assuming that HCl is limiting.

Because I'm given both amounts of the reactants,

I have to do the calculation twice.

So I start with my 10 grams of HCl,

and I use my molar mass,

36.46 grams per mole of HCl.

And then I look at my mole to mole ratio.

In this case, I'm going to have 2 moles of HCl for every 1 mole of H2S.

And then again I use my molar mass of H2S, 34.08 grams per mole of H2S.

And now I'm going to be able to calculate the amount of H2S that could be

produced if HCl is my limiting reagent.

And what I find is that it's 4.67 grams of H2S.

So I see that the iron sulfide is my limiting reagent.

This is the maximum amount of product that I can

produce because I run out of the iron sulfide before I consume all of the HCl.

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