2. Stoichiometry

3. Stoichiometry is a section of chemistry that involves using relationships between reactants and products in a chemical reaction to determine desired quantitative data In a balanced reaction, both sides of the equation have the same number of elements. The stoichiometric coefficient is the number written in front of atoms, ion and molecules to balance the number of each element on both sides of the equation 6CO 2 (g) + 6H 2 O(l)  1C 6 H 12 O 6 (s) + 6O 2 (g)

4. The most common stoichiometric problem will present you with a certain amount of a reactant and then ask how much of a product can be formed. Here is a generic chemical equation: 3 A + 4 B  2 D + 1 F

5. Steps for solving Stoichiometry Problems 1.Balance the chemical equation 2.Convert the given mass to moles by using molar mass or the substance 3.Construct a molar proportion between the substances you are looking for 4.Using the molar mass of the unkown substance, convert the moles just calculated to mass.

6. Stoichiometry Road Map aA + bB cC + dD Mass A Atoms Molecules A Mol AMol B Mass B Atoms Molecules B Mol Ratio Molar mass 6.022 x 10 23 Molar mass 6.022 x 10 23

7. 3 A + 4 B  2 D + 1 F How many moles of F are produced from 1.00 mol of A? 1 mol A 3 mol A 1 mol F = 0.333 mol F How many moles of D are produced from 5.00 mol of B? 5 mol B 4 mol B 2 mol D = 2.50 mol D

8. Working a Stoichiometry Problem 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 1. Identify reactants and products and write the balanced equation. Al+O2O2 Al 2 O 3 b. What are the reactants? a. Every reaction needs a yield sign! c. What are the products? d. What are the balanced coefficients? 43 2

9. = 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 2 x 101.96 ÷ 26.98 ÷ 4 = 12.3 g Al 2 O 3 1. What is your starting point? 6.50 g of Aluminum 2. What is your ending point?g of aluminum oxide Al+O2O2 Al 2 O 3 43 2

10. How many moles of lithium carbonate are produced when 5.3 mol CO 2 are reacted? CO 2 (g) + 2LiOH(s)  Li 2 CO 3 (s) + H 2 O(l) 1. What is your starting point? 2. What is your ending point? 5.3 mol of CO 2 mol of Li 2 CO 3 5.3 mol CO 2 1 mol CO 2 1 mol Li 2 CO 3 = 5.3 mol Li 2 CO 3

11. 3 A + 4 B  2 D + 1 F How many grams of F are produced from 1.00 mol of A? If MM of F is 10.0g/mol. 1 mol A 3 mol A 1 mol F = 3.33g F How many grams of D are produced from 5.00 mol of B? MM of D is 20.0g/mol 5 mol B 4 mol B 2 mol D = 50.0g D 1 mol F 10 g F 1 mol D 20 g D

12. What is the mass of glucose (C 6 H 12 O 6 ) produced from 3.00 mol of water (H 2 O)? 6CO 2 (g) + 6H 2 O(l)  C 6 H 12 O 6 (s) + 6O 2 (g) 3 mol H 2 O 1. What is your starting point? 2. What is your ending point? 3.00 mol of H 2 O g of C 6 H 12 O 6 6 mol H 2 O 1 mol C 6 H 12 O 6 =90.1 g C 6 H 12 O 6 180.81g C 6 H 12 O 6

13. 6CO 2 (g) + 6H 2 O(l)  C 6 H 12 O 6 (s) + 6O 2 (g) What is the mass of oxygen (O 2 ) produced from 2.50 mol of water (H 2 O)? 1. What is your starting point? 2. What is your ending point? 2.50 mol of H 2 O g of O 2 2.5 mol H 2 O 6 mol H 2 O 6 mol O 2 1 mol O 2 32.0 g O 2 =80.0 g O 2

14. 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) How many moles of NO are formed from 824 g of NH 3 ? 1. What is your starting point? 2. What is your ending point? 824 g of NH 3 mol of NO 824 g NH 3 17.03g 1 mol NH 3 4 mol NH 3 4 mol NO = 48.4 mol NO

15. 3 A + 4 B  2 D + 1 F How many grams of F are produced from 5.00g of A? If MM of F is 10.0g/mol and MM of A is 25.0g/mol 1 mole A 3 mole A 1 mole F = 0.677g F How many grams of D are produced from 5.00g of B? MM of D is 20.0g/mol and MM of B is 10.0g/mol 1 mole B 4 mole B 2 mole D =5.00g D 1 mole F 10 g F 1 mole D 20 g D 5 g A 25 g A 5 g B 10 g B

16. Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) How many grams of SnF 2 are produced from the reaction of 30.00 g HF? 1. What is your starting point? 2. What is your ending point? 30.00 g of HF g SnF 2 1 mol HF 2 mol HF 1 mol SnF 2 = 117.5g SnF 2 1 mol SnF 2 156.71 g SnF 2 30.00g HF 20.01g HF

17. Real World Application STOICHIOMETRY Think of a balanced chemical equation as a recipe; the number of moles of each particle input (reactants) indicates the amount of ingredients needed to prepare the food, and the number of moles of each particle output (products) indicates how much of each type of final food one will end up with. The type of input that will be used up first determines the maximum amount of possible output.

18. Click Below for the Video Lectures Stoichiometry