1. Calculations with Limiting Reagents Limiting reactant – reactant (reagent) that is used up first and thus limits how much product can be made The other reactant is said to be in excess

2. 2 eggs + 1 bag chips  50 cookies 1.Stoichiometric quantities = combine 2 eggs and 1 bag of chips, make 50 cookies 2.Have 4 eggs and 1 bag of chips: Limiting reagent: chip Reagent in excess: eggs Can make _____ cookies Answer: 50 cookies

3. N 2 + 3 H 2  2 NH 3 5.00 grams of N 2 and 2.50 grams of H 2 are combined and allowed to react. 1.Which reactant is limiting and which is in excess? –Limiting: N 2 (results in 6.07 g NH 3 ) –Excess: H 2 (would result in 14.0 g NH 3 if all reacted) 2.How many grams of NH 3 will be made? –6.07 g NH 3

4. Zn + 2 AgNO 3  2 Ag + Zn(NO 3 ) 2 A piece of zinc weighing 2.00 grams is placed in a solution containing 2.50 grams of silver nitrate. How many grams of silver (Ag) will form? –First need to determine which reagent is limiting and which is in excess –Limiting: AgNO 3  1.59 g Ag –Excess Zn

5. Percent Yield The amount of product collected after a reaction is often less than the maximum amount possible. Why might this occur?

6. Percent Yield Percent yield is a measure of what % of the maximum yield is actually made or collected. –Measure of efficiency of the reaction or process % Yield = actual yield x 100% theoretical yield

7. Zn + 2 AgNO 3  2 Ag + Zn(NO 3 ) 2 A piece of zinc weighing 2.00 grams is placed in a solution containing 2.50 grams of silver nitrate. Upon completion of the reaction 1.43 grams of silver are collected. What is the percent yield for this reaction? –Based on slide 4…. %yield= 1.43 g/1.59 g x 100% = 89.9%

8. 4 NH 3 + 5 O 2  4 NO + 6 H 2 O 1.How many grams of NO are formed when 1.50 grams of NH 3 reacts with 1.00 grams of O 2 ? Which reactant is limiting and which is in excess? Limiting O 2  make 0.750 g NO Excess: NH 3 – would make 2.65 g NO if all could react How much of the reactant in excess will be left over after the reaction is complete? Leftover NH 3 = 1.50 g -.425 g => 1.08 g

9. CH 4: Types of Reactions, Solution Stoichiometry Water as solvent Electrolytes Solution composition – Molarity Molarity calculations Reactions in solution Types of reactions - lab

10. Water as solvent Water can dissolve: – small polar molecules Water molecules are attracted to the polar molecules –many ionic compounds Water molecules hydrate the ions – this separates the ions and allows them to move freely through the water

11. Water is a polar molecule w ith areas of positive and negative charge H H O  +  -

12. Hydrogen bonding in water Polar molecules can H bond to the water molecules (see board).

13. Ionic Bonding

14. Water Hydrates Ions

15. Electrolytes Electrolyte – substance that conducts electricity when dissolved in water –To conduct electricity there must be charged particles free to move –This occurs when an ionic compund dissolves in water.

16. Types of Electrolytes Strong electrolytes – substances that ionize completely in water –Soluble salts –Strong acids HCl, HNO 3, H 2 SO 4 –Strong bases Group IA hydroxides - NaOH, KOH

17. Types of Electrolytes Weak electrolytes – substances that ionize slightly in water –Few of the dissolved particles form ions –Weak acids All acids but: HCl, HNO 3, H 2 SO 4 HCN, HC 2 H 3 O 2 –Weak bases Often have N in their formula NH 3

18. Types of Electrolytes Nonelectrolytes – substances that dissolve in water*, but do not form ions –Polar molecular compounds Alcohols Sugars * others define as any substance that does not conduct electricity

19. Concentration Unit - Molarity Molarity = moles solute L solution

20. Molarity Calculations 1.Molarity of a given solution 2.How to make a specific volume of solution of specific M 3.Calculations using M 4.Stoichiometry in aqueous solution

21. Making Solutions Describe how to make 1.00 L of a 2.00 M NaOH solution. Describe how to make 500. mL of a 2.00M NaOH solution. Describe how to make 375 mL of a 0.150M Pb(NO 3 ) 2 solution.

22. Molarity of Solutions What is the molarity of a solution made by dissolving 17.00 g of NaOH in enough water to make 250. mL of solution? What is the molarity of a solution made by dissolving 5.00 g NaCl in enough water to make 75.0 mL of solution?

23. From Molarity to Moles Molarity links moles and volume. If you know 2 of these you can calculate the 3 rd. We can use molarity as a conversion factor or take the algebraic approach. M = mol L

24. Molarity Calculations How many moles of HCl are in 125 mL of 3.00 M HCl? What volume of 1.50 M H 2 SO 4 contains 0.325 moles of H 2 SO 4 ?

25. Reaction Stoichiometry Mg + 2 HCl  MgCl 2 + H 2 How many mL of 6.0 M HCl are required to fully react 0.40 grams of Mg?

26. Reaction Stoichiometry NaOH + HCl  NaCl + H 2 O It takes 13.0 mL of 0.200 M NaOH to react 5.00 mL of HCl solution. What is the concentration of the HCl solution?

27. Reaction Stoichiometry H 2 SO 4 + 2 NaOH  products It takes 23.0 mL of 0.200 M NaOH to react 7.00 mL of H 2 SO 4 solution. What is the concentration of the H 2 SO 4 solution?