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Stoichiometry Chapter 12
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Stoichiometry Chapter 12

Feb 07, 2016

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Page 1: Stoichiometry Chapter 12

StoichiometryChapter 12

Page 2: Stoichiometry Chapter 12

Stoichiometry • STOY-KEE-AHM-EH-TREE – Founded by Jeremias Richter, a German chemist– Greek orgin

• stoikheion – element & metron – measure

• The branch of chemistry that deals with the quantities of substances that enter into, and are produced by, chemical reactions.

• Provides the quantitative relationship between reactants and products in a chemical reaction.

Page 3: Stoichiometry Chapter 12

Balanced Chemical Equations• Coefficients are used to balance chemical

equations. • BCE illustrate that chemical reactions follow

Law of Conservation of Mass– The number and types of atoms on the reactants

side is equal to the number and types of atoms on the products.

The coefficients are molar ratios of reactants and products.We can use the molar ratios to convert from one substance to another using Stoichiometry

Page 4: Stoichiometry Chapter 12

Example• What is the mass of Al2O3 that can be made

from 25 grams of aluminum?• Balanced Equation: 4 Al + 3 O2 2 Al2O3

BCE

Page 5: Stoichiometry Chapter 12

1. You have to use units to understand the process2. Units include g, mol, L, particles & chemical formulas3. Convert your given to moles4. You have to use units to understand the process5. Use BCE coefficients to convert from one substance to another6. You have to use units to understand the process7. Convert answer to desired unit 8. You have to use units to understand the process9. Units include g, mol, L, particles & chemical formulas10. It really is not that hard

You Should Know About Stoichiometry

Page 6: Stoichiometry Chapter 12

Percent Yield• Indicates the percent of theoretical yield that was obtained

from the final product in an experiment.

• The closer to 100% the better

• Percent Yield = actual mass x 100 theoretical mass

Page 7: Stoichiometry Chapter 12

Percent Error

• Percent Error = (actual mass – theoretical mass) x 100

theoretical mass • The lower the percent error the better

Page 8: Stoichiometry Chapter 12

Limiting Reactants• The limiting reactant is the reactant present in the

smallest stoichiometric amount– In other words, it’s the reactant you’ll run out of first

5N2 and 9H2 6NH3 and 2N2

Hydrogen is limiting

Hydrogen is limiting

Page 9: Stoichiometry Chapter 12

How 2 Calculate the Limiting Reactant

• Convert each of the given reactants to a product• The reactant that produces the least product is the limiting

reactant.Example:• Nickel replaces silver from silver nitrate in solution according to the following equation:

2AgNO3 + Ni → 2Ag + Ni(NO3)2 If you have 22.9 g of Ni and 112 g of AgNO3 ,what mass of nickel(II) nitrate would be produced?

• 22.9g Ni x 1 mol Ni x 1 mol Ni(NO3)2 x 182.7g Ni(NO3)2 = 71.3 g Ni(NO3)2

58.7g Ni 1 mol Ni 1 mol Ni(NO3)2

• 112 g AgNO3 x 1 mol AgNO3 x 1 mol Ni(NO3)2 x 182.7g Ni(NO3)2 = 60.2 g Ni(NO3)2

169.9g AgNO3 2 mol AgNO3 1 mol Ni(NO3)2

• AgNO3 is limiting & 60.2 g Ni(NO3)2 can be produced

Page 10: Stoichiometry Chapter 12

Limiting Reactant Challenge• Given 20 mL HCl (1.46g HCl) calculate the amount of

Mg to add to generate the maximum production of Hydrogen gas with Mg still limiting the reaction.

• Calculate the percent yield of Hydrogen gas