Explore
For Enterprise
Join for Free
Log In

Arts and Humanities
Business
Computer Science
Data Science
Information Technology
Health
Math and Logic
Personal Development
Physical Science and Engineering
Social Sciences
Language Learning
Degrees and Certificates

Explore all of Coursera

Browse
Search
For Enterprise
Log In
Join for Free

5.02b How many grams of magnesium do we need

Loading...

Chemistry

University of Kentucky

4.7 (639 ratings) | 65K Students Enrolled

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.

Reviews

4.7 (639 ratings)

5 stars
478 ratings
4 stars
132 ratings
3 stars
14 ratings
2 stars
8 ratings
1 star
7 ratings

TM

Jun 21, 2017

This is a great course! Very well explained! It's managing to keep my attention on 100% and that's not an easy task for anybody. Thank you for sharing your knowledge with us!

M

Dec 30, 2016

I enjoyed every minute of this course and I have learnt so much. Thank you Coursera for this opportunity. I'll definitely recommend this course to my friends and colleagues.

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

Taught By

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

How many grams of magnesium do we need to start with to produce 95 grams of

magnesium nitride?

How many grams of nitrogen do we need?

Well the first thing I notice when I look at this problem is that there are no

coefficients in my equation.

And that's usually a really good sign that I at least need to check to make sure I

have a balanced equation.

And what I see is that my nitrogens are balanced cause I

have two on either side but my magnesiums are not.

So I'm actually going to need to add a coefficient here,

add a three in there, in order to balance the magnesium.

So now I have a balanced chemical equation.

The next thing I'm going to do is actually put in my molar masses, just so

when I start to do my calculations I already have this information compiled.

Remember that nitrogen is N2.

So I have to put 28.02 grams per mole, and

then magnesium nitrite is 100.93 grams per mole.

So it's asking first about the amount of magnesium I need to

produce some amount of magnesium nitride.

So I'm going to start with my 95.00 grams of Mg3N2 and

I'm going to convert to moles using the molar mass of magnesium nitride.

And then I'm going to use my mole to mole ratio.

So I'm talking about magnesiums, so I notice that I have

one mole of magnesium nitride for every three moles of magnesium.

And then I can use the molar mass of magnesium,

24.31 grams per mole of magnesium to find the grams.

Now a couple things to note here, one is that I'm going to cancel my grams of

magnesium nitride, moles of magnesium nitride,

moles of magnesium and I'm going to be left with grams of magnesium.

The other thing to note is that I started with the amount of product.

And it really doesn't matter which substance I start with and

which substance I'm going to.

Remember that those stoichiometric coefficients from our balanced chemical

equation will apply regardless of whether I'm looking at a reactant in the product,

two reactants, two products or

if I'm given information about the product and asking information about the reactant.

It does not matter as long as I use the coefficients for

my balanced chemical equation.

So now I need to do the calculation, and

what I find is 68.65 grams of magnesium are needed

in order to produce the 95 grams of magnesium nitride.

Now I need to do a similar calculation going from

the magnesium nitride to the amount of nitrogen I need.

So here we have, again, 95 grams of magnesium nitride.

And I'm going to say use my molar mass here,

100.93 grams per mole of magnesium nitride.

And now again I have to have a mole ratio.

This time it's one mole of magnesium nitride to one mole of nitrogen.

And then I'm going to use my molar mass of nitrogen,

which is 28.02 grams per mole of nitrogen.

Note that my units cancel out, my grams of magnesium nitride, moles of

magnesium nitride, moles of nitrogen, and I'm left with grams of nitrogen.

And I find that I need 26.37 grams of nitrogen.

Now this is not a limiting reagent problem because I

wasn't given two amounts of reactant.

I was given a single product.

And so what we've actually calculated here are the stoichiometric amounts.

If I were to mix exactly 68.65 grams of magnesium with 26.37 grams of nitrogen,

what I would find is that both of the reactants would be completely consumed.

It would not have an excess or a limiting reagent, because both of these

amounts were calculated based on producing the 95 grams of magnesium nitride.

Explore our Catalog

Join for free and get personalized recommendations, updates and offers.

Coursera provides universal access to the world’s best education, partnering with top universities and organizations to offer courses online.

© 2019 Coursera Inc. All rights reserved.

Download on the App Store

Get it on Google Play

Coursera
About
Leadership
Careers
Catalog
Certificates
MasterTrack™ Certificates
Degrees
For Enterprise
For Government

Community
Learners
Partners
Developers
Beta Testers
Translators

Connect
Blog
Facebook
LinkedIn
Twitter
Instagram
YouTube
Tech Blog

More
Terms
Privacy
Help
Accessibility
Press
Contact
Directory
Affiliates