Presentation is loading. Please wait.

Presentation is loading. Please wait.

Stoichiometry Chemical Reactions and Quantities Mole Relationships in Chemical Equations Mole-Mole Mole-Grams Grams-Grams Theoretical Yield % Yield Limiting.

Similar presentations


Presentation on theme: "Stoichiometry Chemical Reactions and Quantities Mole Relationships in Chemical Equations Mole-Mole Mole-Grams Grams-Grams Theoretical Yield % Yield Limiting."— Presentation transcript:

1 Stoichiometry Chemical Reactions and Quantities Mole Relationships in Chemical Equations Mole-Mole Mole-Grams Grams-Grams Theoretical Yield % Yield Limiting Reactants

2 Mole-Mole Factors Write and balance the equation Show the mole-to-mole ratio between two of the substances in a balanced equation Use the coefficients of two substances in the balanced equation

3 Writing Mole Factors 4 Fe + 3 O 2 2 Fe 2 O 3 Fe and O 2 4 mole Fe and 3 mole O 2 3 mole O 2 4 mole Fe Fe and Fe 2 O 3 4 mole Feand2 mole Fe 2 O 3 2 mole Fe 2 O 3 4 mole Fe O 2 and Fe 2 O 3 3 mole O 2 and 2 mole Fe 2 O 3 2 mole Fe 2 O 3 3 mole O 2

4 Mole Factor 3 H 2 (g) + N 2 (g) 2 NH 3 (g) A. A mole factor for H 2 and N 2 is 1) 3 mole N 2 2) 1 mole N 2 3) 1 mole N 2 1 mole H 2 3 mole H 2 2 mole H 2 B. A mole factor for NH 3 and H 2 is 1) 1 mole H 2 2) 2 mole NH 3 3) 3 mole N 2 2 mole NH 3 3 mole H 2 2 mole NH 3

5 Solution 3 H 2 (g) + N 2 (g) 2 NH 3 (g) A. A mole factor for H 2 and N 2 is 2) 1 mole N 2 3 mole H 2 B. A mole factor for NH 3 and H 2 is 2) 2 mole NH 3 3 mole H 2

6 Chemical Calculations 4 Fe + 3 O 2 2 Fe 2 O 3 How many moles of Fe 2 O 3 are produced when 6.0 moles O 2 react?

7 Chemical Calculations 4 Fe + 3 O 2 2 Fe 2 O 3 How many moles of Fe 2 O 3 are produced when 6.0 moles O 2 react? 6.0 mole O 2 x 2 mole Fe 2 O 3 = 4.0 mole Fe 2 O 3 3 mole O 2

8 Mole Ratio Calculations 4 Fe + 3 O 2 2 Fe 2 O 3 How many moles of Fe are needed to react with 12.0 mole of O 2 ? 1) 3.00 mole Fe 2) 9.00 mole Fe 3) 16.0 mole Fe

9 Solution 4 Fe + 3 O 2 2 Fe 2 O mole O 2 x 4 mole Fe = 16.0 mole Fe 3 mole O 2

10 Mole Ratio/Grams Calculations 4 Fe + 3 O 2 2 Fe 2 O 3 How many grams of O 2 are needed to produce mole of Fe 2 O 3 ? 1) 38.4 g O 2 2) 19.2 g O 2 3) 1.90 g O 2

11 Solution 2) 19.2 g O 2

12 Solution Combined 2) 19.2 g O mole Fe 2 O 3 x 3 mole O 2 x 32.0 g O 2 2 mole Fe 2 O 3 1 mole O 2 = 19.2 g O 2

13 Mass of A Reaction The reaction between H 2 and O 2 produces 13.1 g of water. How many grams of O 2 reacted? Write the equation H 2 (g) + O 2 (g)H 2 O (g)

14 Mass of A Reaction The reaction between H 2 and O 2 produces 13.1 g of water. How many grams of O 2 reacted? Write the equation H 2 (g) + O 2 (g)H 2 O (g) Balance the equation

15 Mass of A Reaction The reaction between H 2 and O 2 produces 13.1 g of water. How many grams of O 2 reacted? Write the equation H 2 (g) + O 2 (g)H 2 O (g) Balance the equation 2 H 2 (g) + O 2 (g)2 H 2 O (g)

16 Organize data mole bridge 2 H 2 (g) + O 2 (g)2 H 2 O (g) ? g 13.1 g Plan g H 2 O mole H 2 O mole O 2 O 2 Setup

17 Organize data mole bridge 2 H 2 (g) + O 2 (g)2 H 2 O (g) ? g 13.1 g Plan g H 2 O mole H 2 O mole O 2 O 2 Setup 13.1 g H 2 O x 1 mole H 2 O x 1 mole O 2 x 32.0 g O g H 2 O2 mole H 2 O 1 mole O 2 = 11.6 g O 2

18 Points to Remember 1. Read an equation in moles 2. Convert given amount to moles 3.Use mole factor to give desired moles 4. Convert moles to grams grams (given)grams (desired) moles (given)moles (desired)

19 Stoichiometric Calculations How many grams of O 2 are need to react 50.0 grams of Na in the reaction 4 Na + O 2 2 Na 2 O Complete the set up: 50.0 g Na x 1 mole Na x ________ x _______ 23.0 g Na

20 Solution 4 Na + O 2 2 Na 2 O 50.0 g Na x 1 mole Na x 1 mole O 2 x 32.0 g 23.0 g Na 4 mole Na 1 mole O 2 = 17.4 g O 2

21 Stoichiometric Calculations Acetylene gas C 2 H 2 burns in the oxyactylene torch for welding. How many grams of C 2 H 2 are burned if the reaction produces 75.0 g of CO 2 ? C 2 H 2 + O 2 CO 2 + H 2 O

22 Solution 2 C 2 H O 2 4 CO H 2 O 75.0 g CO 2 x 1 mole CO 2 x 2 mole C 2 H 2 x 26.0 g C 2 H g CO 2 4 mole CO 2 1 mole C 2 H 2 = 22.2 g C 2 H 2

23 Stoichiometry Part 2 Calculations with Equations Limiting Reactions Percent Yield

24 Stoichiometry How many O 2 molecules will react with 505 grams of Na to form Na 2 O? 4 Na + O 2 2 Na 2 O Complete the set up: 505 g Na x 1 mol Na x ________ x _______ 23.0 g Na

25 Solution 4 Na + O 2 2 Na 2 O 505 g Na x 1 mol Na x 1 mol O 2 x 6.02 x g Na 4 mol Na 1 mol O 2 = 3.30 x molelcules

26 Limiting Reactants If the amounts of two reactants are given, the reactant used up first determines the amount of product formed.

27 So, you want to work for a sandwich shop? Suppose you are preparing cheese sandwiches. Each sandwich requires 2 pieces of bread and 1 slice of cheese. If you have 4 slices of cheese and 10 pieces of bread, how many cheese sandwiches can you make?

28 Making Cheese Sandwiches Sandwich = Sandwich =

29 Sandwich Stoichiometry How many sandwiches can you make? ____ slices of bread + ____ slices of cheese = ____ sandwiches What is left over? ________________ What is the limiting reactant?

30 Solution How many sandwiches can you make? __10__ slices of bread + __4__ slices of cheese = __4__ sandwiches What is left over? _2 slices of bread What is the limiting reactant? cheese

31 Hints for Limiting Reactant Problems 1. For each reactant amount given, calculate the moles (or grams) of a product it could produce. 2.The reactant that produces the smaller amount of product is the limiting reactant. 3. The number of moles of product produced by the limiting reactant is ALL the product possible. There is no more limiting reactant left.

32 Percent Yield You prepared cookie dough to make 5 dozen cookies. The phone rings while a sheet of 12 cookies is baking. You talk too long and the cookies burn. You throw them out (or give them to your dog.) The rest of the cookies are okay. How many cookies could you have made (theoretical yield)? How many cookies did you actually make to eat? (Actual yield)

33 Key Phrases Actual yield is the amount of product actually recovered from an experiment Theoretical (possible) yield is the maximum amount of product that could be produced from the reactant. Percent Yield is the actual yield compared to the maximum (theoretical yield) possible.

34 Calculating Percent Yield What is the percent yield of cookies? Percent Yield = Actual Yield (g) recovered X 100 Possible Yield (g) % cookie yield = 48 cookies x 100 = 80% yield 60 cookies


Download ppt "Stoichiometry Chemical Reactions and Quantities Mole Relationships in Chemical Equations Mole-Mole Mole-Grams Grams-Grams Theoretical Yield % Yield Limiting."

Similar presentations


Ads by Google