5.05b A student needs 625 g of zinc sulfide

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來自 University of Kentucky 的課程

Chemistry

508 個評分

University of Kentucky

Chemistry

508 個評分

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.

從本節課中

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

與講師見面

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

A student needs 625 grams of zinc sulfide, a white pigment, for an art project.

He can synthesize it using the reaction above.

How many grams of zinc nitrate will he need if he can make the zinc

sulfide in 85% yield?

Assume that he as plenty of sodium sulfide.

So in this problem, we've got lots of information that's given.

And the first thing we want to do is kind of summarize that information.

And so I'm going to start by seeing the amount of the product that we need,

the 625 grams of the zinc sulfide, okay?

He can synthesize it using the reaction above.

And it's in an 85% yield.

And we have plenty of sodium sulfide.

So I'm just going to put an XS there to stand for excess.

Okay?

So we have plenty of that.

So really we're only worried about how much zinc nitrate do we need?

So, I've got a couple of things going on here.

I've got to figure out how much zinc sulfide do I need to plan on making,

if I know I'm only going to have 85% yield.

And then I need to use stoichiometry to figure out how much zinc nitrate is

going to be needed to make that amount of zinc sulfide.

So let's first look at what we have for the zinc sulfide in that percent yield.

Remember that percent yield equals the actual,

or the experimental amount, over the theoretical, times 100.

So I can use that equation to actually figure out how much I

need to plan on making.

So I know my percent yield is 85%.

My actual yield needs to be 625,

because this is what I actually need to end up with at the end of the experiment,

and what I'm actually solving for here is the theoretical amount.

How much do I need to plan on making so that I end up with 625 grams?

So we know that x is going to be greater than 625.

And when I do the calculation, I'm going to have to solve for that x.

And I find that x is going to be equal to 735 grams.

So that's how much I need to plan on making so that I end up with that 625,

because my reaction is not going to go to completion.

Now, a couple of things before we continue on with this problem that we

want to take note of.

One, we know the amount of zinc sulfide we need to plan on making.

The other thing we need to notice is this ratio.

For every one mole of zinc sulfide,

I need to have actually reacted one mole of zinc nitrate.

So we're going to bring over the information here that we're going to

need to solve the rest of this problem.

And we said we needed to plan on making 735 grams of the zinc sulfide.

Remember that it was a one to one ratio between the zinc sulfide and

the zinc nitrate.

And that came from our balanced chemical equation.

And I also know that I have molar masses for

zinc sulfide, which is 97.47 grams per mole.

And for zinc nitrate, the molar mass is 189.36 grams per mole.

So with this information I can actually set up my stoichiometry calculation.

Notice that the amount, the known amount is the 735 grams of zinc sulfide,

and I'm going to have to get that to moles,

because remember my ratio, that one to one ratio, is in terms of moles.

So I use the molar mass, and I'm going to put the grams on the bottom because I

need those grams to cancel with grams.

Then I'm going to put in my mole ratio, that one mole of

zinc sulfide to one mole of zinc nitrate.

And even though it's a one to one ratio and it's not changing my

actual calculation any, I'm still going to include it just to get in the habit and so

that all my units cancel out correctly.

Now I have one mole of zinc nitrate.

Now I can use my molar mass of 189.36 grams

per mole of zinc nitrate to get to the grams of zinc nitrate.

Before I do the calculation, I'm going to cancel out grams and grams.

Moles of zinc sulfide, moles of zinc sulfide,

moles of zinc nitrate with moles of zinc nitrate.

And what I'm left with are grams of zinc nitrate,

which is what I'm actually trying to find in the problem.

And when I do the calculation, I end up with 1428 grams

of zinc nitrate, and to get that in the correct number of significant figures,

I'm going to have to write that in scientific notation.

So it would be 1.43 times ten to the third grams of zinc nitrate.

So that's the amount of zinc nitrate I need to start with, so that

in the end, I'm produce my 625 grams of the zinc sulfide.

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