Reaction Stoichiometry I Introduction to Reaction Stoichiometry A

Reaction Stoichiometry

I. Introduction to Reaction Stoichiometry A. Definitions 1. Stoichiometry - deals with the amount of reactants and products in chemical reactions 2. Mole Ratio - conversion factor that relates the number of moles of any two substances involved in a chemical reaction

B. Types of Stoichiometry Problems (A is the given, B is the unknown) 1. mole-mole 2. mole-mass 3. mass-mole 4. mass-mass (mole A (mass A mole B) mass B)

C. Diagram A is the given B is what you need to find - the unknown *All problems will include a conversion factor 1 mole A # mole B mole A molar mass B MASS A -----> MOLE A ------> MOLE B ------> MASS B molar mass A # mole A 1 mole B

III. Mole Ratios and Coefficients in Equations A. 2 H 2 + O 2 -----> 2 H 2 O 1) H 2, O 2 2) H 2, H 2 O 2 mol H 2 1 mol O 2 2 mol H 2 O 2 mol H 2 3) O 2, H 2 O 1 mol O 2 2 mol H 2 O 1 mol O 2 B. 2 KCl. O 3 -----> 2 KCl + 3 O 2 1) KCl. O 3, KCl 2)KCl. O 3 , O 2 3) KCl, O 2

II. Solving Stoichiometry Problems (All Problems Include Mole Ratios) A. Mole-Mole electricity 1. 2 Na + Cl 2 -----> 2 Na. Cl A – Given B - find a) 4 moles of Na will react with (? ) moles of Cl 2? 4. 0 mol Na X 1. 0 mol Cl 2 = 2 mol Cl 2 1 2. 0 mol Na

A. Mole –Mole Problem 1. 2 Na + Cl 2 -----> 2 Na. Cl A - Given B - Find b)6 moles of Na will form (? ) moles of Na. Cl? 6 mol Na X 2 mol Na. Cl = 6 mol Na. Cl 1 2 mol Na

B. Mole-Mass 1. Ca. O + H 2 O -----> Ca(OH)2 A B a)2 moles of Ca. O will produce(? )grams of Ca(OH)2 2 mol Ca. O x 1 mol Ca(OH)2 x 74. 1 g Ca(OH)2 = 1 1 mol Ca. O 1 mol Ca(OH)2 148 g Ca(OH) 2

1. Ca. O + H 2 O -----> Ca(OH)2 b) How many grams of Ca. O are needed to form 2 moles of Ca(OH)2? 2 mol Ca(OH)2 x 1 mol Ca. O x 56. 1 g Ca. O = 1 1 mol Ca(OH)2 1 mol Ca. O 112 g Ca. O

C. Mass-Mole 1. Zn + 2 HCl -----> Zn. Cl 2 + 2 H 2 a) How many moles of zinc chloride are formed when 196. 2 grams of zinc react with hydrochloric acid? 196. 2 g Zn x 1 mol Zn. Cl 2 = 3. 001 mol Zn. Cl 2 1 65. 38 g Zn

1. Zn + 2 HCl -----> Zn. Cl 2 + 2 H 2 b) How many moles of zinc are necessary to form 67. 70 grams of zinc chloride? 67. 70 g Zn. Cl x 1 mol Zn. Cl 2 x 1 mol Zn = 1 136. 29 g Zn. Cl 2 1 mol Zn. Cl 2. 4967 mol Zn

D. Mass-Mass 1. CH 4 + 2 O 2 -----> CO 2 + 2 H 2 O a) How many grams of carbon dioxide are formed when 64. 0 grams of oxygen react with methane? 64. 0 g O 2 x 1 mol CO 2 x 44. 0 g CO 2 = 1 32. 0 g O 2 2 mol O 2 1 mol CO 2 44. 0 g CO 2

1. CH 4 + 2 O 2 -----> CO 2 + 2 H 2 O b) How many grams of methane are needed to produce 48. 0 grams of water? 48. 0 g H 2 O x 1 mol CH 4 x 16. 0 g CH 4 = 1 18. 0 g H 2 O 2 mol H 2 O 1 mol CH 4 21. 3 g CH 4

III. Limiting Reactant A. Definitions 1. limiting reactant – limits the extent of reaction and determine the amount of product (reactant that is used up first) 2. excess reactant – portion of reactant that remains after a reaction is complete

B. What are some examples that model limiting reactant and excess reactant? 1. hamburger + bun 2. cake recipe requires 2 eggs and 3 cups of flour, and 1 cup sugar

C. How is the Limiting Reactant Determined? 1. Divide number of moles of each reactant by its coefficient in the balanced equation 2. The substance with the smaller number is the limiting reactant. 3. The substance with the larger number is the excess reactant.

D. Example Determine the limiting reactant, the excess reactant and the number of moles of each product? (6 moles of H 2 combined with 3 moles of N 2) 1. 3 H 2(g) 6 mol + N 2(g) 2 NH 3(g) 3 mol a) 6/3 = 2 H 2 is excess 3/1 = 1 N 2 is limiting b) 3 mol N 2 x 2 mol NH 3 = 1 1 mol N 2 6 mol NH 3

Determine the limiting reactant, the excess reactant and the number of moles of each product 2. Zn(s) + 2 HCl(aq) Zn. Cl 2(aq) + H 2(g) 2. 5 mol Zn +3. 5 mol HCl are combined a) 2. 5/1 = 2. 5 Zn is excess 3. 5/2 = 1. 75 HCl is limiting b) 3. 5 mol HCl x 1 mol Zn. Cl 2 = 1. 8 mol Zn. Cl 2 1 2 mol HCl c) 3. 5 mol HCl x 1 mol H 2 = 1 2 mol HCl 1. 8 mol H 2

IV. Percent Yield A. Definitions 1. theoretical yield – maximum amount of product that can be produced from a given amount of reactant 2. actual yield – amount of product actually produced during a reaction carried out in the lab 3. percent yield – ratio of actual yield to theoretical yield expressed as a percent

% Yield = Actual yield (experiment results) X 100% Theoretical yield (stoichiometric calculations) B. Procedure 1. Determine theoretical yield ( mass stoichiometric calculation) 2. Divide actual yield (experiment results in grams) by theoretical yield 3. Multiply by 100% to get percent yield

1. Zinc reacts with iodine in a synthesis reaction. a. Determine theoretical yield if 125. 0 g of zinc is used. b. Determine the % yield if the mass of the product zinc iodide recovered is 556 g. Zn(s) + I 2(s) Zn. I 2(s) a)125. 0 g Zn x 1 mol Zn. I 2 x 329. 2 g Zn. I 2 = 707. 4 g 1 65. 4 g Zn 1 mol Zn Theoretical Yield is 707. 4 g Zn. I 2 b) Actual yield…………. . 556 = Theoretical yield……. 707. 4 78. 6% (percent yield)

2. Water decomposes when subject to an electric current to form hydrogen gas and oxygen gas. In the lab, 60. 0 g of oxygen is recovered when 72. 0 g of water is decomposed. What is theoretical yield? What is the percent yield? electricity 2 H 2 O 2 H 2 + O 2 72. 0 g H 2 O x 1 mol O 2 x 32. 0 g O 2 = 64. 0 g O 2 1 18. 0 g H 2 O 2 mol H 2 O 1 mol O 2 Actual Yield………… 60. 0 g Theoretical Yield…. . . 64. 0 g = 93. 8% (percent yield)