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5.06e How many grams of LiCl do we need

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Chemistry

University of Kentucky

4.7 (700 ratings) | 69K 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.

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Jun 21, 2017

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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

Transcript

How many grams of lithium chloride do we need to add in order to prepare 450 millilitres of a 1.4 molar solution? So a couple of things here, one, we know the substance we're dealing with. We've got lithium chloride. And, remember that when we talk about molarity, what we really mean is for this will be 1.4 moles per litre, so moles of lithium chloride per litre of solution. So, thinking about molarity in terms of this expanded unit can help us figure out how we're going to set this problem up. So, let's first come up with a plan of how we're going to get to our answer. We know we want grams of lithium chloride at the end, and if I want to get to grams, I know I can get there from moles of lithium chloride because I can figure out what the molar mass is for that particular substance. And so, I've got to look here and figure out how am I going to get to moles of lithium chloride. Well, notice that I'm given this relationship between the moles and the volume of lithium chloride in the terms of the molarity. And so, if I have the volume of my lithium chloride. Then I can get two moles of lithium chloride, which can get me two grams of lithium chloride. So, the first we want to do is actually find the molar mass of our lithium chloride. So, lithium has a mass of 6.94, chlorine 35.45. And we get a molar mass of 42.39 grams per mole. And so, I'm going to be able to use that later in my problem to get to the final answer. So the first thing I'm going to start with, is my 450 millilitres. And I'm going to have to convert that to liters because I notice that molarity is in moles per liter. So, 1,000 millilitres in 1 liter. Then, I'm going to use my molarity to get from my volume to my moles. So, 1.4 moles of lithium chloride per liter. Now, I have moles on top. I'm going to put moles on the bottom and grams on top here. 42.39 grams per mole, because again I want to make sure my units are cancelling out. And that's what I'm actually going to do now, is go back through and see that millilitres cancels with millilitres, liters cancel, moles cancel, and what I'm left with are units of grams. And that's what I'm looking for in the problem. And when I do the math, what I end up with is 26.7 grams of lithium chloride. But I need to round this to two significant digits, so I'll have 27 grams of lithium chloride.

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