Chemistry - Chp 12 - Stoichiometry - PowerPoint

Chapter 12 “Stoichiometry” Mr. Mole

Section 12.1 The Arithmetic of Equations ,[object Object],[object Object]

Let’s make some Cookies! ,[object Object],[object Object],[object Object]

Stoichiometry is… ,[object Object],[object Object],[object Object],[object Object]

#1. In terms of Particles ,[object Object],[object Object],[object Object]

Example: 2H 2 + O 2 -> 2H 2 O ,[object Object],[object Object],2 formula units Al 2 O 3 form 4 atoms Al and 3 molecules O 2 Now read this: 2Na + 2H 2 O  2NaOH + H 2

#2. In terms of Moles ,[object Object],[object Object],[object Object],[object Object]

#3. In terms of Mass ,[object Object],[object Object],[object Object],2 moles H 2 2.02 g H 2 1 mole H 2 = 4.04 g H 2 1 mole O 2 32.00 g O 2 1 mole O 2 = 32.00 g O 2 36.04 g H 2 + O 2 36.04 g H 2 + O 2 + reactants

In terms of Mass (for products) ,[object Object],2 moles H 2 O 18.02 g H 2 O 1 mole H 2 O = 36.04 g H 2 O 36.04 g H 2 + O 2 = 36.04 g H 2 O The mass of the reactants must equal the mass of the products. 36.04 grams reactant = 36.04 grams product

#4. In terms of Volume ,[object Object],[object Object],[object Object],[object Object],67.2 Liters of reactant ≠ 44.8 Liters of product!

Practice: ,[object Object],[object Object]

Section 12.2 Chemical Calculations ,[object Object],[object Object]

Mole to Mole conversions ,[object Object],[object Object],2 moles Al 2 O 3 3 mole O 2 or 2 moles Al 2 O 3 3 mole O 2 These are the two possible conversion factors to use in the solution of the problem.

Mole to Mole conversions ,[object Object],[object Object],3.34 mol Al 2 O 3 2 mol Al 2 O 3 3 mol O 2 = 5.01 mol O 2 If you know the amount of ANY chemical in the reaction, you can find the amount of ALL the other chemicals! Conversion factor from balanced equation

Practice: ,[object Object],[object Object],( 9.6 mol ) ,[object Object],( 8.95 mol ) ,[object Object],( 4.94 mol )

Steps to Calculate Stoichiometric Problems ,[object Object],[object Object],[object Object],[object Object],[object Object]

Mass-Mass Problem: 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 4Al + 3O 2  2Al 2 O 3 = 6.50 g Al ? g Al 2 O 3 1 mol Al 26.98 g Al 4 mol Al 2 mol Al 2 O 3 1 mol Al 2 O 3 101.96 g Al 2 O 3 (6.50 x 1 x 2 x 101.96) ÷ (26.98 x 4 x 1) = 12.3 g Al 2 O 3 are formed

Another example: ,[object Object],[object Object],Answer = 17.2 g Cu

Volume-Volume Calculations: ,[object Object],[object Object],17.5 L O 2 22.4 L O 2 1 mol O 2 2 mol O 2 1 mol CH 4 1 mol CH 4 22.4 L CH 4 = 8.75 L CH 4 Notice anything relating these two steps? 22.4 L O 2 1 mol O 2 1 mol CH 4 22.4 L CH 4

Avogadro told us: ,[object Object],[object Object],[object Object],1 mole = 22.4 L @ STP

Shortcut for Volume-Volume? ,[object Object],[object Object],17.5 L O 2 2 L O 2 1 L CH 4 = 8.75 L CH 4 Note : This only works for Volume-Volume problems.

Section 12.3 Limiting Reagent & Percent Yield ,[object Object],[object Object]

“ Limiting” Reagent ,[object Object],[object Object],[object Object],[object Object]

Limiting Reagents - Combustion

How do you find out which is limited? ,[object Object],[object Object],[object Object]

[object Object],[object Object],10.6 g Cu 63.55g Cu 1 mol Cu 2 mol Cu 1 mol Cu 2 S 1 mol Cu 2 S 159.16 g Cu 2 S = 13.3 g Cu 2 S 3.83 g S 32.06g S 1 mol S 1 mol S 1 mol Cu 2 S 1 mol Cu 2 S 159.16 g Cu 2 S = 19.0 g Cu 2 S = 13.3 g Cu 2 S Cu is the Limiting Reagent, since it produced less product.

Another example: ,[object Object],[object Object],[object Object],[object Object],Excess = 4.47 grams

The Concept of: A little different type of yield than you had in Driver’s Education class.

What is Yield? ,[object Object],[object Object],[object Object],[object Object],[object Object],x 100

Example: ,[object Object],[object Object],[object Object],[object Object],[object Object],= 6.78 g Cu = 13.8 g Cu = 49.1 %

Details on Yield ,[object Object],[object Object],[object Object],[object Object]

Chemistry - Chp 12 - Stoichiometry - PowerPoint

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