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Chemistry 103 Lecture 20
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Chemistry 103 Lecture 20. Chemical Calculations A mixture of 25.0g of H 2 and an excess of N 2 react according to the following equation: 3H 2 (g) + N.

Dec 19, 2015

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Page 1: Chemistry 103 Lecture 20. Chemical Calculations A mixture of 25.0g of H 2 and an excess of N 2 react according to the following equation: 3H 2 (g) + N.

Chemistry 103

Lecture 20

Page 2: Chemistry 103 Lecture 20. Chemical Calculations A mixture of 25.0g of H 2 and an excess of N 2 react according to the following equation: 3H 2 (g) + N.

Chemical Calculations A mixture of 25.0g of H2 and an excess of N2 react

according to the following equation:

3H2(g) + N2(g) -----> 2NH3(g)

How many liters of NH3 are produced at STP?

Page 3: Chemistry 103 Lecture 20. Chemical Calculations A mixture of 25.0g of H 2 and an excess of N 2 react according to the following equation: 3H 2 (g) + N.

Chemical Calculations A mixture of 25.0g of H2 and an excess of N2 react

according to the following equation:

3H2(g) + N2(g) -----> 2NH3(g)

How many liters of NH3 are produced at STP 745mmHg and 23oC?

Page 4: Chemistry 103 Lecture 20. Chemical Calculations A mixture of 25.0g of H 2 and an excess of N 2 react according to the following equation: 3H 2 (g) + N.

Outline

I. Solutions - vocabulary - Molarity defined - Solution Stoichiometry

Page 5: Chemistry 103 Lecture 20. Chemical Calculations A mixture of 25.0g of H 2 and an excess of N 2 react according to the following equation: 3H 2 (g) + N.

SOLUTIONS - Chapter 12

Page 6: Chemistry 103 Lecture 20. Chemical Calculations A mixture of 25.0g of H 2 and an excess of N 2 react according to the following equation: 3H 2 (g) + N.

Matter

Pure Substance Mixture

Element Compound Heterogeneous Homogeneous

Page 7: Chemistry 103 Lecture 20. Chemical Calculations A mixture of 25.0g of H 2 and an excess of N 2 react according to the following equation: 3H 2 (g) + N.

Matter

Pure Substance Mixture

Element Compound Heterogeneous Homogeneous

(Solutions)

Page 8: Chemistry 103 Lecture 20. Chemical Calculations A mixture of 25.0g of H 2 and an excess of N 2 react according to the following equation: 3H 2 (g) + N.

Mixtures

Homogeneous Mixtures (solutions) The composition is uniform throughout The different parts of the mixture are not visible air, salt water, alloys (i.e. brass, bronze)

Heterogeneous Mixtures The composition of substances is not uniform The composition varies from one part of the mixture to another The different parts of the mixture are visible potting soil, oil and water

Page 9: Chemistry 103 Lecture 20. Chemical Calculations A mixture of 25.0g of H 2 and an excess of N 2 react according to the following equation: 3H 2 (g) + N.

Solution vocabulary

Solute - that which is dispersed in a solution (present in lesser amount)

Reacting Component!!!!! Solvent - dispersing agent (present in

greatest amount)

Page 10: Chemistry 103 Lecture 20. Chemical Calculations A mixture of 25.0g of H 2 and an excess of N 2 react according to the following equation: 3H 2 (g) + N.

Solutes Spread evenly

throughout the solution.

Are not visible, but can give a color to the solution.

Nature of Solutes in Solutions

Copyright © 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings

Page 11: Chemistry 103 Lecture 20. Chemical Calculations A mixture of 25.0g of H 2 and an excess of N 2 react according to the following equation: 3H 2 (g) + N.

Formation of a Solution

Na+ and Cl- ions• On the surface of a

NaCl crystal are attracted to polar water molecules.

• In solution are hydrated as several H2O molecules surround each.

Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings

Page 12: Chemistry 103 Lecture 20. Chemical Calculations A mixture of 25.0g of H 2 and an excess of N 2 react according to the following equation: 3H 2 (g) + N.

Solution vocabulary - UnsaturatedUnsaturated solutions Contain less than the

maximum amount of solute.

Can dissolve more solute. Dissolved

solute

Copyright © 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings

Page 13: Chemistry 103 Lecture 20. Chemical Calculations A mixture of 25.0g of H 2 and an excess of N 2 react according to the following equation: 3H 2 (g) + N.

Solution vocabulary - SaturatedSaturated solutions Contain the

maximum amount of solute that can dissolve.

Have undissolved solute at the bottom of the container.

Copyright © 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings

Page 14: Chemistry 103 Lecture 20. Chemical Calculations A mixture of 25.0g of H 2 and an excess of N 2 react according to the following equation: 3H 2 (g) + N.

Concentration Terms

Concentration Designates the amount of solute dissolved in a given

quantity of solution

There are many ways of designating concentration. Percent concentration

Percent by mass (m/m) Percent by volume (v/v)

Molarity = M

Page 15: Chemistry 103 Lecture 20. Chemical Calculations A mixture of 25.0g of H 2 and an excess of N 2 react according to the following equation: 3H 2 (g) + N.

Molarity

Molarity (M) A Conversion Factor between the amount of solute

and the volume of solution!!!

M = Molarity = moles of solute

L of solution

Page 16: Chemistry 103 Lecture 20. Chemical Calculations A mixture of 25.0g of H 2 and an excess of N 2 react according to the following equation: 3H 2 (g) + N.

Making a Solution Steps for making a solution:

1. Weigh out the solute, put it in a volumetric flask.

2. Add some water, mix the solute and the solvent.

3. Add water to the line on the volumetric flask.

Page 17: Chemistry 103 Lecture 20. Chemical Calculations A mixture of 25.0g of H 2 and an excess of N 2 react according to the following equation: 3H 2 (g) + N.

Molarity Problems

Calculate the Molarity of 73.0g of HCl in 2.00L of solution.

Calculate the grams of solute needed to prepare 325mL of a 0.100M CaCl2 solution.

Page 18: Chemistry 103 Lecture 20. Chemical Calculations A mixture of 25.0g of H 2 and an excess of N 2 react according to the following equation: 3H 2 (g) + N.

Molarity in Chemical Reactions In a chemical reaction, The volume and molarity of a solution are used to determine

the moles of a reactant or product.

M ( mol ) x volume (L) = moles solute L

If molarity, M (mol/L), and moles are given, the volume (L) can be determined.moles x 1 L = volume (L) solution moles

Page 19: Chemistry 103 Lecture 20. Chemical Calculations A mixture of 25.0g of H 2 and an excess of N 2 react according to the following equation: 3H 2 (g) + N.

Summary of Chemical Reactions

Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings

Page 20: Chemistry 103 Lecture 20. Chemical Calculations A mixture of 25.0g of H 2 and an excess of N 2 react according to the following equation: 3H 2 (g) + N.

Using Molarity of ReactantsHow many mL of 3.00 M HCl solution are needed to

completely react with 4.85 g CaCO3?2HCl(aq) + CaCO3(s) CaCl2(aq) + CO2(g) + H2O(l)

Page 21: Chemistry 103 Lecture 20. Chemical Calculations A mixture of 25.0g of H 2 and an excess of N 2 react according to the following equation: 3H 2 (g) + N.

Learning Check

How many mL of a 0.150 M Na2S solution are needed to completely react 18.5 mL of 0.225 M NiCl2 solution?

NiCl2(aq) + Na2S(aq) NiS(s) + 2NaCl(aq)

Page 22: Chemistry 103 Lecture 20. Chemical Calculations A mixture of 25.0g of H 2 and an excess of N 2 react according to the following equation: 3H 2 (g) + N.

Learning Check

What volume, in liters, of H2 gas measured at STP can be produced from 50.0mL of 6.0M H2SO4 solution and 25.0g of Ni according to the reaction:

Ni(s) + H2SO4(aq) ------> NiSO4(aq) + H2(g)