Presentation is loading. Please wait.

Presentation is loading. Please wait.

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.

Similar presentations


Presentation on theme: "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."— Presentation transcript:

1 Chemistry 103 Lecture 20

2 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 2 (g) -----> 2NH 3 (g) How many liters of NH 3 are produced at STP?

3 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 2 (g) -----> 2NH 3 (g) How many liters of NH 3 are produced at STP 745mmHg and 23 o C?

4 Outline I. Solutions - vocabulary - Molarity defined - Solution Stoichiometry

5 SOLUTIONS - Chapter 12

6 Matter Pure Substance Mixture Element Compound Heterogeneous Homogeneous

7 Matter Pure Substance Mixture Element Compound Heterogeneous Homogeneous (Solutions)

8 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

9 Solution vocabulary Solute - that which is dispersed in a solution (present in lesser amount) Reacting Component!!!!! Solvent - dispersing agent (present in greatest amount)

10 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

11 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 H 2 O molecules surround each. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

12 Solution vocabulary - Unsaturated Unsaturated 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

13 Solution vocabulary - Saturated Saturated 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

14 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

15 Molarity Molarity (M)  A Conversion Factor between the amount of solute and the volume of solution!!! M = Molarity = moles of solute L of solution

16 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.

17 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 CaCl 2 solution.

18 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

19 Summary of Chemical Reactions Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

20 Using Molarity of Reactants How many mL of 3.00 M HCl solution are needed to completely react with 4.85 g CaCO 3 ? 2HCl(aq) + CaCO 3 (s) CaCl 2 (aq) + CO 2 (g) + H 2 O(l)

21 Learning Check How many mL of a 0.150 M Na 2 S solution are needed to completely react 18.5 mL of 0.225 M NiCl 2 solution? NiCl 2 (aq) + Na 2 S(aq) NiS(s) + 2NaCl(aq)

22 Learning Check What volume, in liters, of H 2 gas measured at STP can be produced from 50.0mL of 6.0M H 2 SO 4 solution and 25.0g of Ni according to the reaction: Ni(s) + H 2 SO 4 (aq) ------> NiSO 4 (aq) + H 2 (g)


Download ppt "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."

Similar presentations


Ads by Google