1. Stoichiometry

2. Definition Reaction Stoichiometry is the calculation of quantitative (measurable) relationships of the reactants and products in chemical reactions Reaction Stoichiometry is the calculation of quantitative (measurable) relationships of the reactants and products in chemical reactionscalculation quantitative reactantsproductschemical reactionscalculation quantitative reactantsproductschemical reactions

3. 4 types of stoichiometry problems 1. Mole – mole 2. Mole – mass 3. Mass – mole 4. Mass - mass

4. Reaction Stoichiometry Road Map

5. Stoichiometry: Mole to Mole ALWAYS ALWAYS start with a balanced chemical equation Given: Al 2 O 3  Al + O 2 Balanced Equation: 2Al 2 O 3  4Al + 3O 2 Therefore, two moles of aluminum oxide can produce four moles of aluminum and three moles of oxygen Q: If six moles of aluminum oxide are used, how many moles of aluminum are produced?

6. Problem Set-up 2Al 2 O 3  4Al + 3O 2 Start with the number in the question. Start with the number in the question. 6 moles Al 2 O 3 6 moles Al 2 O 3 Set-up a factor label problem, and move the units to the bottom to cancel Set-up a factor label problem, and move the units to the bottom to cancel 6 moles Al 2 O 3 X 4 moles Al 6 moles Al 2 O 3 X 4 moles Al 2 moles Al 2 O 3 2 moles Al 2 O 3 Complete the math, and add labels to your final answer. Complete the math, and add labels to your final answer. 24 = 12 moles of Al 24 = 12 moles of Al2

7. YOU TRY Balance your equation left to right: CO 2 + LiOH  Li 2 CO 3 + H 2 O Balance your equation left to right: CO 2 + LiOH  Li 2 CO 3 + H 2 O CO 2 + 2LiOH  Li 2 CO 3 + H 2 O CO 2 + 2LiOH  Li 2 CO 3 + H 2 O How many moles of LiOH are required to react with 30 mol of CO 2 ? How many moles of LiOH are required to react with 30 mol of CO 2 ? 60 moles of LiOH 60 moles of LiOH

8. Stoichiometry: Mole to Mass Given 3.00 mol of water and an excess of carbon dioxide how many grams of glucose (C 6 H 12 O 6 ) will be produced? CO 2 + H 2 O  C 6 H 12 O 6 + O 2 6 CO 2 + 6 H 2 O  C 6 H 12 O 6 + 6 O 2 3.00 mol H 2 O * (1 mol C 6 H 12 O 6 / 6 mol H 2 O) * (180g C 6 H 12 O 6 /1 mol C 6 H 12 O 6 ) = 90.0 g C 6 H 12 O 6 3.00 mol H 2 O * (1 mol C 6 H 12 O 6 / 6 mol H 2 O) * (180g C 6 H 12 O 6 /1 mol C 6 H 12 O 6 ) = 90.0 g C 6 H 12 O 6

9. Stoichiometry: Mass to Mole If 8.00 g of SO 2 reacts with an excess of carbon how many moles of CS 2 are formed? If 8.00 g of SO 2 reacts with an excess of carbon how many moles of CS 2 are formed? C + SO 2  CS 2 + CO Balanced Equation: 5 C + 2 SO 2  CS 2 + 4 CO Balanced Equation: 5 C + 2 SO 2  CS 2 + 4 CO 8.00 g SO 2 * (1 mol SO 2 /64.1 g SO 2 ) * 8.00 g SO 2 * (1 mol SO 2 /64.1 g SO 2 ) * (1 mol CS 2 /2 mol SO 2 ) = 0.0624 mol CS 2

10. Stoichiometry: Mass to Mass Fluorine reacts with sodium chloride to form sodium fluoride and chlorine. Write the balanced equation for this single-displacement reaction. F 2 + 2NaCl  2NaF + Cl 2 F 2 + 2NaCl  2NaF + Cl 2 How many grams of sodium fluoride can be produced if 45 grams of fluorine are consumed? How many grams of sodium fluoride can be produced if 45 grams of fluorine are consumed? 45 g F 2 * (1 mol F 2 /38 g F 2 ) * (2 mol NaF/1 mol F 2 ) * (42 g NaF/1 mol NaF) 45 g F 2 * (1 mol F 2 /38 g F 2 ) * (2 mol NaF/1 mol F 2 ) * (42 g NaF/1 mol NaF)

11. Limiting reactant Reactant that limits the amount of the other reactants used and the amount of product formed Reactant that limits the amount of the other reactants used and the amount of product formed

12. Limiting Reactant

13. Everyday Example Nestle Toll House® Chocolate Chip Cookie Recipe Nestle Toll House® Chocolate Chip Cookie Recipe 2 1/4 cups all-purpose flour 2 1/4 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon baking soda 1 teaspoon salt 1 teaspoon salt 1 cup (2 sticks, 1/2 pound) butter, softened 1 cup (2 sticks, 1/2 pound) butter, softened 3/4 cup granulated [white] sugar 3/4 cup granulated [white] sugar 3/4 cup packed brown sugar 3/4 cup packed brown sugar 1 teaspoon vanilla extract 1 teaspoon vanilla extract 2 eggs 2 eggs 2 cups (12-ounce package) NESTLE TOLL HOUSE Semi- Sweet Chocolate Morsels 2 cups (12-ounce package) NESTLE TOLL HOUSE Semi- Sweet Chocolate Morsels 1 cup chopped nuts 1 cup chopped nuts If you have 40 ounces of semi- sweet chocolate morsels, 6 cups of chopped nuts, 4.5 teaspoons of salt, and 1.5 cups of brown sugar how many batches of cookies can you bake with each ingredient? 40 ounces of morsels = 3 batches 6 cups of nuts = 6 batches 4.5 teaspoons of salt = 4 batches 1.5 cups sugar = 2 batches You can only make two batches of cookies. After two batches you will run short on needed ingredients.

14. Everyday Example #2 The Hollidaysburg Area chemistry club decide to prepare fruit baskets as a fund raiser. A complete fruit basket contains six oranges, two bananas, one pineapple, and three grapefruit. The Hollidaysburg Area chemistry club decide to prepare fruit baskets as a fund raiser. A complete fruit basket contains six oranges, two bananas, one pineapple, and three grapefruit. The secretary of the club placed the following order with out consulting their chief of stoichiometric affairs: 5 dozen oranges, 2 dozen bananas, 1 dozen pineapples, and 2 dozen grapefruits. The secretary of the club placed the following order with out consulting their chief of stoichiometric affairs: 5 dozen oranges, 2 dozen bananas, 1 dozen pineapples, and 2 dozen grapefruits.

15. Everyday Example #2 Needed: 6 oranges, 2 bananas, 1 pineapple, and 3 grapefruit Needed: 6 oranges, 2 bananas, 1 pineapple, and 3 grapefruit Ordered: 5 dozen oranges, 2 dozen bananas, 1 dozen pineapples, and 2 dozen grapefruits Ordered: 5 dozen oranges, 2 dozen bananas, 1 dozen pineapples, and 2 dozen grapefruits Write a balanced equation showing the correct ratios relating the reactants and product in the preparation of a fruit basket. Write a balanced equation showing the correct ratios relating the reactants and product in the preparation of a fruit basket. 6O + 2B + 1P + 3G  1 Basket 6O + 2B + 1P + 3G  1 Basket

16. Everyday Example #2 How many complete fruit baskets can be prepared? How many complete fruit baskets can be prepared? Ordered: 5 dozen oranges, 2 dozen bananas, 1 dozen pineapples, and 2 dozen grapefruits Ordered: 5 dozen oranges, 2 dozen bananas, 1 dozen pineapples, and 2 dozen grapefruits 6O + 2B + 1P + 3G  1 Basket 6O + 2B + 1P + 3G  1 Basket Example Calculation: Example Calculation: 5 dozen oranges = 60 oranges 5 dozen oranges = 60 oranges 60 oranges x 1 basket = 10 baskets 60 oranges x 1 basket = 10 baskets 6 oranges 6 oranges Bananas = 12 baskets Bananas = 12 baskets Pineapples = 12 baskets Pineapples = 12 baskets Grapefruit = 8 baskets Grapefruit = 8 baskets

17. Everyday Example #2 What fruit is the limiting reagent? What fruit is the limiting reagent? GRAPEFRUIT GRAPEFRUIT How much of the other fruits are left over? How much of the other fruits are left over? Example Calculation Example Calculation 8 baskets x 6 oranges = 48 oranges 8 baskets x 6 oranges = 48 oranges 1 basket 1 basket 60 oranges ordered – 48 oranges used = 12 oranges left 60 oranges ordered – 48 oranges used = 12 oranges left 8 bananas 8 bananas 4 pineapples 4 pineapples

18. Chemical Example Silicon dioxide (quartz) reacts with hydrogen fluoride according to the following reaction: Silicon dioxide (quartz) reacts with hydrogen fluoride according to the following reaction: SiO 2 (s) + 4HF(g)  SiF 4 (g) + 2H 2 O(l) SiO 2 (s) + 4HF(g)  SiF 4 (g) + 2H 2 O(l)

19. Example SiO 2 (s) + 4HF(g)  SiF 4 (g) + 2H 2 O(l) If 2.0 mol of HF is combined with 4.5 mol of SiO 2, which is the limiting reactant? If 2.0 mol of HF is combined with 4.5 mol of SiO 2, which is the limiting reactant? 2.0 mol HF x [1 mol SiF 4 / 4 mol HF] = 2.0 mol HF x [1 mol SiF 4 / 4 mol HF] = 0.50 mol SiF 4 0.50 mol SiF 4 4.5 mol SiO 2 x [1 mol SiF 4 /1 SiO 2 ] = 4.5 mol SiO 2 x [1 mol SiF 4 /1 SiO 2 ] = 4.5 mol SiF 4 The HF is the reagent that limits the amount of product that can be made, it is the LIMITING REAGENT! The HF is the reagent that limits the amount of product that can be made, it is the LIMITING REAGENT!

20. Example 2 CH 4 + 2O 2  CO 2 + 2H 2 O If 32 grams of methane and 32 grams of oxygen react, which is the limiting reagent? If 32 grams of methane and 32 grams of oxygen react, which is the limiting reagent? 32 g CH 4 x [1 mol CH 4 /16 g CH 4 ] x 32 g CH 4 x [1 mol CH 4 /16 g CH 4 ] x [1 mol CO 2 /1 mol CH 4 ] = 2 mol CO 2 32 g O 2 x [1 mol O 2 /32 g O 2 ] x 32 g O 2 x [1 mol O 2 /32 g O 2 ] x [1mol CO 2 /2 mol O 2 ] = 0.5 mol CO 2 Since O 2 limits the amount of product that can be made it is the limiting reagent. Since O 2 limits the amount of product that can be made it is the limiting reagent.

21. Percent Yield Percent Yield= Percent Yield=[Actual Yield/Theoretical Yield] * 100% C 6 H 6 + Cl 2  C 6 H 5 Cl + HCl g of C 6 H 6 reacts with an excess of Cl 2 the actual yield is 38.8 g of C 6 H 5 Cl. What is the percent yield of this reaction? If 36.8 g of C 6 H 6 reacts with an excess of Cl 2 the actual yield is 38.8 g of C 6 H 5 Cl. What is the percent yield of this reaction? 36.8 g C 6 H 6 * [1 mol C 6 H 6 /78.1 g C 6 H 6 ] * 36.8 g C 6 H 6 * [1 mol C 6 H 6 /78.1 g C 6 H 6 ] * [1 mol C 6 H 5 Cl/1 mol C 6 H 6 ] * [113 g C 6 H 5 Cl/1 mol C 6 H 5 Cl] = 53.2 g C 6 H 5 Cl [38.8 g C 6 H 5 Cl/53.2 g C 6 H 5 Cl] * 100% = 72.9% [38.8 g C 6 H 5 Cl/53.2 g C 6 H 5 Cl] * 100% = 72.9%