12+Stoichiometry

=→Chapter 12 Stoichiometry=

12.1 The arithmetic of Equations
 * Objectives:**
 * How is a balanced equation like a recipe?
 * How do chemists use balanced chemical equations?
 * In terms of what quantities can you interpret a balanced chemical equation?
 * What quantities are conserved in every chemical reaction?


 * How do you write a word equation?
 * How do you write a skeleton equation?
 * What are the steps in writing a balanced equation?

12.2 Chemical Calculations
 * How are mole ratios used in chemical calculations?
 * What is the general procedure for solving a stoichiometric problem?
 * What is the general procedure for solving a stoichiometric problem?

12.3 Limiting Reagent and Percent Yield
 * How is the amound of product in a reaction affected by an insufficient quantity of any of the reactants?
 * What does the percent yield of a reaction measure?

=Outlined= =Notes= 12.1 The Arithmetic of Equations I Using Everyday Equations 1. A balanced chemical equation provides the samw kind of quantitative information that a recipe does. II Using Balanced Chemical Equations 1. Chemists use balanced chemical equations as a bases to calculate how much reactant is needed or product is formed in a reaction. 2. Stoichiometry- The calculation of quantitites in a chemical reactions is a subject of chemistry. III Interpreting Chemicsl Equations 1. A b The reactant of which the balanced chemical equation can be interpreted in terms of different quantities, including numbers of atoms, molecules, or moles;mass;and volume. A. Number of Atoms a) The Atomic level, a balanced equation indicates that the number and type of each atom that makes up each reactantalso makes up each product. B. Number of Molecules b) The balanced equation indicates that one molecule of nitrogen reacts with three molecules of hydrogen.

II Chemical Calculations A. Writing and Using Mole Ratios 1. In chemical calculations, mole ratios are used to convert between moles of reactant and moles of product, between moles of reactants, or between moles of products. a. __mole ratio__ - a conversion factor derived from the coefficients of a balanced chemical equation interpreted in terms of moles. 2.If a given sample is measured in grams, the mass can be converted to moles by using the molar mass. Then the mole ratio from the balanced equation can be used to calculate the number of moles of the unknown. a.Mass-mass problems are solved in basically the same way as mole-mole problems. B. Other Stoichiometric Calculations 1. In a typical stoichiometric problem, the given quantity is first converted to moles. Then the mole ratio from the balanced equation is used to calculate the number of moles of the wanted substance. Finally, the moles are converted to any other unit of measurement related to the unit mole, as the problem requires. a. In a volume-volume problem, the 22.4L/mol factors cancel out. = = = 12.3 = = = III Limitin Reagent and Percent Yield

A Limiting and Excess Reagents 1 In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms. 2 __Limiting Reactant-__ the reactant of which there s not enough; when completely consumed up, the reaction stops 3 Example To produce two moles of water you, would need (2X2g) 4g of H2 and 32g of O2. If you had only 2g of hydrogen only one mole of water would form. 4 __Excess Reactants-__ Other reactants, present in quantities great than those needed to react with the quantity of limiting reagent

B Percent Yield 1 Theoretical Yield- The amount of product that would result if all the limiting reagent reacted 2 Actual Yield- The quantity of products that are the experimental results from a reaction in the lab 3 Percent Yield- The ratio of the actual yield to the theoretical yield expressed as a percent 4 The percent yield is a measure of the efficiency of a reaction carried out in the laboratory  __Ex __  13.1g CaO is produced Actual Yield 24.8g CaCO3 at start CaCO3 à  CaO + CO2 What is the percent yield? __24.8g CaCO3 | 1 mol CaCO2 | 1 mol CaO | 56.1g CaO. __ | 100.1 g CaCo3 | 1 mol CaCO3| 1 mol CaO = 13.9 g CaO

__Percent Yield __= __. actual .__ = __13.1g CaO__ X 100% = 94% Theoretical 13.9g CaC

=Reference Pages= = 12.3 =







à Practice Problems [] 1a) How many moles of chlorine gas (Cl2) would react with 5 moles of sodium (Na) according to the following chemical equation? (Balance equation.) 2 Na + Cl2 --> 2 NaCl 2.5 mol Cl2 1b) Using the equation (after it is balanced) above, determine the amount of product that can be produced from 24.7 g Na. 62 g NaCl 1c) How many molecules of product would be produced from 24.7g Na? 6.5 E23 NaCl molecules

**__ Stoichiometry example problems: __**
-**Mole to Mole problem:** How many moles of oxygen are necessary to react completely with four moles of propane?(assume that the numbers behind the elements are small, and that there are lines in between the problem) C3H8 + 5O2-->3CO2 +4H2O __**4 moles C3H8 5 mole O2**__ Answer to the question: __20 moles O2__ -**Volume to Volume** problem**:** If 30 mL of hydroen are produced, how many millileters of oxygen are produced? 2H2O --> 2H2 + O2 __**30 Ml H2 1 mole H2 1 mole O2 22.4 O2**__ Answer to problem: __15 mL__ -**Mass to Mass problem:** How many grams of potassium chloride are produced if 25 grams of potassium chloride decompose? 2KCLO3 --> 2KCL + 3O2 __**25 grams KCLO3 1 mole KCLO3 2 moles KCL 74.6 KCL**__ Answer: __15 grams KCL__ -**Mixed Mole problem:**What volume of NH3 at STP is produced if 25.0g of N2 is reacted with an excess of H2? N2 + 3H2 --> 2NH3 · **Volume-Volume:** What volume of nitrogen is necessary to react with 5.0 liters of hydrogen to produce ammonia? (Assume STP N2 + 3H2 2NH3  __ 5.0 L H2 1 mol H2 1 mol N2 22.4 L N2 __ Answer: __1.7 L N2__  22.4 L H2 3 mol H2 1 mol N2 ·  **Volume**** -Volume: ** What volume of ammonia was produced in the problem above? (Still assume STP)  __1.7 L N2 1 mol N2 2 mol NH3 22.4 L NH3__ Answer: __3.4 L NH3__  22.4 L N2 1 mol N2 1 mol NH3 ** If 2.8 grams of sodium were used for the reaction, and an excess of iodine was present, how many grams of sodium iodide would be produced? ** 2.8g Na || 1 mole Na || 1 mole NaI || 150 g NaI || = 18g NaI How many grams of hydrogen are necessary to react completely with 50.0g of nitrogen in the below reaction? N2 + 3H2 → 2NH3 How many grams of silver chloride are produced from 5.0g of silver nitrate reacting with an excess of barium chloride? 2AgNO3 + BaCl2 → 2AgCl + Ba(NO3)2
 * 1 mole C3H8**
 * 22.4 L H2 2 moles H2 1 mole O2**
 * 122.6 g KCLO3 2 mol KCLO3 1 mol KCL**
 * __25.0 g N2 1 mol N2 2 mol NH3 22.4 L NH3__** Answer: __40.0L NH3__
 * 28g N2 1 mol N2 1 mol NH3**
 * 1. Na + I2 ---> NaI **
 * || 23g Na || 1 mole Na || 1 mole NaI ||
 * Mass-Mass Problems**
 * __50.0g N2 l 1 mol N2 l 3 mol H2 l 2g H2__ || = 10.7 g H2 ||  ||   ||
 * l 28g N2 l 1 mol N2 l 1 mol H2 ||  ||   ||   ||

__5.0g AgNO3 l 1 mol AgNO3 l 2 mol AgCl l 143.4g AgCl__ = 2.581g AgCl l 277.8g AgNO3 l 2 mol AgNO3 l 1 mol AgCl

Mole-Mole C3H8 + 5O2 → 3CO2 + 4H2O How many moles of oxygen are necessary to react completely with four moles of propane(C3H8)?

__4 mol C3H8 | 5 mol O2__ = 20 mol O2
 * 1 mol C3H8

Mixed Mole 2KClO3 → 2KCl + 3O2 If 5.0 g of KClo3 is decomposed, what volume of O2 is produced at STP? __5.0g KClO3 | 1 mol KClO3 | 3 mol O2 |22.4 L O2__ = 1.4 L O2
 * 122.6g KClO3 | 2 mol KClO3 | 1 mol O2

1. N2 + 3H2 à <span style="font-size: 12pt; line-height: 115%; font-family: 'Times New Roman','serif';"> 2NH3 How many grams of NH3 can be produced from the reaction of 28g of N2 and 25g of H2?
 * LIMITING REAGENT PROBLEMS:**

__28g N2 │1 mol N2 │2 mol NH3│17g NH3__ = 34g NH3 │ 28g N2 │1 mol N2 │1 mol NH3

__25g H2 │1 mol H2 │2 mol NH3│17g NH3__ = 140g NH3 │ 2g H2 │3 mol H2 │1 mol NH3

=34g NH3 <span style="font-size: 12pt; font-family: 'Times New Roman','serif';">2. <span style="font-size: 12pt; line-height: 115%; font-family: 'Times New Roman','serif';">N2 + 3H2 à <span style="font-size: 12pt; line-height: 115%; font-family: 'Times New Roman','serif';"> 2NH3 How much of the excess reagent in Problem 1 is left over?

__34g NH3 │1 mol NH3 │3 mol H2│29g H2__ = 6g H2 │ 17g NH3 │2 mol NH3 │1 mol H2

25g H2 – 6g H2 =19g H2 <span style="font-size: 12pt; font-family: 'Times New Roman','serif';">3. <span style="font-size: 12pt; line-height: 115%; font-family: 'Times New Roman','serif';">Mg + 2HCl à <span style="font-size: 12pt; line-height: 115%; font-family: 'Times New Roman','serif';"> MgCl2 + H2 What volume of hydrogen at STP is produced from the reaction of 50.0g of Mg and the equivalent of 75g of HCl?

__50.0g Mg │1 mol Mg │1 mol H2│22.4L H2__ = 46.1L H2 │ 24.3g Mg │1 mol Mg │1 mol H2

__75g HCl │1 mol HCl │2 mol H2│22.4 L H2__ = 23L H2 │ 36.5g HCl │2 mol HCl │1 mol H2

=23 L H2 <span style="font-size: 12pt; font-family: 'Times New Roman','serif';">4. <span style="font-size: 12pt; line-height: 115%; font-family: 'Times New Roman','serif';">How much of the excess reagent in problem 3 is left over?

__23L H2 │1 mol H2 │1 mol Mg │24.3g Mg__ = 25g Mg │22.4 L H2 │1 mol H2 │1 mol H2

50.0g Mg – 25g Mg= =25g Mg <span style="font-size: 12pt; font-family: 'Times New Roman','serif';">5. <span style="font-size: 12pt; line-height: 115%; font-family: 'Times New Roman','serif';">3AgNO3 + Na3PO4 à <span style="font-size: 12pt; line-height: 115%; font-family: 'Times New Roman','serif';"> Ag3PO4 + 3NaNO3 Silver nitrate and sodium phosphate are reacted in equal amounts of 200.g each. How many grams of silver phosphate are produced?

__200g AgNO3 │1 mol AgNO3 │1 mol Ag3PO4│419g Ag3PO4__ = 164g Ag3PO4 │170g AgNO3 │3 mol AgNO3│1 mol Ag3PO4

__200g Na3PO4 │1 mol Na3PO4│1 mol Ag3PO4 │419g Ag3PO4__ = 511g Ag3PO4 │164g Na3PO4 │1 mol Na3PO4│1 mol Ag3PO4

=164g Ag3PO4 <span style="font-size: 12pt; font-family: 'Times New Roman','serif';">6. <span style="font-size: 12pt; line-height: 115%; font-family: 'Times New Roman','serif';">How much of the excess reagent in Problem 5 is left? __164g Ag3PO4 │1 mol Ag3PO4 │1 mol Na3PO4 │164g Na3PO4__ = 64.2g Na3PO4 │419g Ag3PO4 │1 mol Ag3PO4│1 mol Na3PO4 200.g Na3PO4 – 64.2g Na3PO4 =

=135.8g Na3PO4





also:
[] [] [] 140g NH3 (3 is subscript) 12.3

A Example How much Aluminum was required for the reaction.

__10.2g Al2 Cl | 1 mol Al2 Cl6 | 2 mol Al | 27.0g Al__ | 266.7 g Al2 Cl6 | 1 mol Al2 Cl6 | 1 mol Al

5.40 g Al given __- 2.07g Al required__

Stoic: Mole-Mole Problems

1. N2 + 3H2 à 2NH3 __2 mol H2 |3 mol H2__ = 6 mol H2
 * 1 mol N2

2. 2KClO3 à 2KCl __6 mol KClO3 | 3 mol O2__ = 9 mol O2 3. Zn + 2HCl à ZnCl2 + H2 __3 mol Zn | 1 mol H2__ = 3 mol H2
 * 1 mol Zn
 * 1 mol Zn

4. C3H8 + 5O2 à 3CO2 + 4H2O __4 mol C3H8 | 5O2__ = 20 mol O2
 * 1 mol C3H8

5. K3PO4 + Al(NO3)3 à 3 KNO3 + AlPO4 __2 mol K3PO4 | 3 KNO3__ = 6 mol KNO3
 * 1 mol K3PO4

Volume to volume How many liters of water are need to make this reaction work? 1. 2H2 + O2 à 2H2O Answer .003L
 * .003 L H2 || 1 mol H2 || 2 mol H2­O || 22.4 L H2O ||  ||
 * || 22.4 L H2 || 2 mol H2 || 1 mol H2O ||  ||

The answer is the same as you start out with because you can cancal out the 22.4 L and the 2 mol.

=3.33g Al left over in excess Assignments=

__Mixed Stoichiometry Problems []__

=Labs= Lab 1- [] Lab 2- [] Lab 3- [] Lab 4- [|**http://www.celinaschools.org/sci/Chem1_stoichiometry.htm**] Lab 5- []

12.3 Lab 1- []

=Sample Test= [] [|stoich quiz] [] [] [] [] [] [] [] [] []http://www.avon-chemistry.com/stoich_prac_test_1.htm = = =**Links**= [] [] [] 12.3 NHH [] [] http://chemistry.about.com/library/weekly/bl061703a.htm [] [] [] [] [] [] [] [] Video Tutor http://www.midnighttutor.com/PercentYieldFull.html [] http://www.science.uwaterloo.ca/~cchieh/cact/c120/stoichio.html[|://www.marietta.edu/~patek/flash/stoich.htmlhttp://wiki.answers.com/Q/How_do_you_solve_stoichiometry_problems] = **http://www.molecularsoft.com/stoichiometry.htm?gclid** =
 * []**
 * CM2p9JzxwZkCFQ9JagodhFQ3vA**=