2. 1 STOICHIOMETRY TUTORIAL www.centralcatholichs.com/faculty/.../conversionsStoichio metry.ppt

3. TEK 8 (E) perform stoichiometric calculations, including determination of mass relationships between reactants and products, calculation of limiting reagents, and percent yield. 8 (E) perform stoichiometric calculations, including determination of mass relationships between reactants and products, calculation of limiting reagents, and percent yield. 2

4. 3 Instructions: This is a work along tutorial. Each time you click the mouse or touch the space bar on your computer, one step of the problem solving occurs. Pressing the PAGE UP key will backup the steps. Get a pencil and paper, a periodic table and a calculator, and let’s get to work.

5. 4 (1-2-3) General Approach For Problem Solving: 1. Clearly identify the Goal or Goals and the UNITS involved. (What are you trying to do?) 2. Determine what is given and the UNITS. 3. Use conversion factors (which are really ratios) and their UNITS to CONVERT what is given into what is desired.

6. 5 Table of Contents: Click on each tab to view problem types. Sample problem 1 Sample problem 2 Converting grams to moles Mole to Mole Conversions View Complete Slide Show Gram-Mole and Gram-Gram Problems Solution Stoichiometry Problems Limiting/Excess/ Reactant and Theoretical Yield Problems :

7. 6 Sample problem for general problem solving. Sam has entered into a 10 mile marathon. Use ALL of the following conversions (ratios) to determine how many inches there are in the race. 5280 ft = 1 mile; 12 inches = 1 ft 1. What is the goal and what units are needed? Goal = ______ inches 2. What is given and its units? 10 miles 3. Convert using factors (ratios). 10 miles = inches633600 Units match Given Goal Convert Menu

8. 7 Sample problem #2 on problem solving. A car is traveling at a speed of 45 miles per hr (45 miles/hr). Determine its speed in kilometers per second using the following conversion factors (ratios). 1 mile = 5280 ft; 1 ft = 12 in; 1 inch = 2.54 cm; k = 1 x 10 3 ; c = 1 x 10 -2 ; 1 hr =60 min; 1 min = 60 s = km s 0.020 Units Match! Goal c cancels c m remains Given This is the same as putting k over k

9. 8 Converting grams to moles. Determine how many moles there are in 5.17 grams of Fe(C 5 H 5 ) 2. Goal = moles Fe(C 5 H 5 ) 2 Given 5.17 g Fe(C 5 H 5 ) 2 Use the molar mass to convert grams to moles. Fe(C 5 H 5 ) 2 2 x 5 x 1.001 = 10.01 2 x 5 x 12.011 = 120.11 1 x 55.85 = 55.85 0.0278 units match

10. 9 Stoichiometry (more working with ratios) Ratios are found within a chemical equation. 2HCl + Ba(OH) 2  2H 2 O + BaCl 2 1 1 2 moles of HCl react with 1 mole of Ba(OH) 2 to form 2 moles of H 2 O and 1 mole of BaCl 2 coefficients give MOLAR RATIOS

11. 10 When N 2 O 5 is heated, it decomposes: 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) a. How many moles of NO 2 can be produced from 4.3 moles of N 2 O 5 ? = moles NO 2 4.3 mol N2O5N2O5 8.6 b. How many moles of O 2 can be produced from 4.3 moles of N 2 O 5 ? = mole O 2 4.3 mol N2O5N2O5 2.2 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 4.3 mol? mol 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 4.3 mol ? mol Mole – Mole Conversions Units match

12. 11 When N 2 O 5 is heated, it decomposes: 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) a. How many moles of N 2 O 5 were used if 210g of NO 2 were produced? = moles N 2 O 5 210 g NO 2 2.28 b. How many grams of N 2 O 5 are needed to produce 75.0 grams of O 2 ? = grams N 2 O 5 75.0 g O2O2 506 gram ↔ mole and gram ↔ gram conversions 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 210g? moles 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 75.0 g ? grams Units match

13. 12 Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? First write a balanced equation. Al(s) + HCl(aq)  AlCl 3 (aq) + H 2 (g) 2 6 2 3 Gram to Gram Conversions

14. 13 Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? Al(s) + HCl(aq)  AlCl 3 (aq) + H 2 (g) 2 6 2 3 Now let’s get organized. Write the information below the substances. 3.45 g ? grams Gram to Gram Conversions

15. 14 Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? Al(s) + HCl(aq)  AlCl 3 (aq) + H 2 (g) 2 6 2 3 3.45 g ? grams Let’s work the problem. = g AlCl 3 3.45 g Al We must always convert to moles.Now use the molar ratio.Now use the molar mass to convert to grams. 17.0 Units match gram to gram conversions

16. 15 Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO 2, is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? b. Determine the limiting reactant. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) First copy down the the BALANCED equation! Now place numerical the information below the compounds.

17. 16 Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO 2, is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? b. Determine the limiting reactant. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 0.15 mol 0.10 mol ? moles Two starting amounts? Where do we start? Hide one

18. 17 Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO 2, is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? b. Determine the limiting reactant. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 0.15 mol 0.10 mol ? moles Hide Based on: KO 2 = mol O 2 0.15 mol KO 2 0.1125

19. 18 Potassium superoxide, KO 2, is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? b. Determine the limiting reactant. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 0.15 mol 0.10 mol ? moles Based on: KO 2 = mol O 2 0.15 mol KO 2 0.1125 Hide Based on: H 2 O = mol O 2 0.10 mol H2OH2O 0.150 Limiting/Excess/ Reactant and Theoretical Yield Problems :

20. 19 Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO 2, is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? Determine the limiting reactant. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 0.15 mol 0.10 mol ? moles Based on: KO 2 = mol O 2 0.15 mol KO 2 0.1125 Based on: H 2 O = mol O 2 0.10 mol H 2 O 0.150 What is the theoretical yield? Hint: Which is the smallest amount? The is based upon the limiting reactant? It was limited by the amount of KO 2. H 2 O = excess (XS) reactant!

21. 20 Theoretical yield vs. Actual yield Suppose the theoretical yield for an experiment was calculated to be 19.5 grams, and the experiment was performed, but only 12.3 grams of product were recovered. Determine the % yield. Theoretical yield = 19.5 g based on limiting reactant Actual yield = 12.3 g experimentally recovered

22. 21 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) If a reaction vessel contains 120.0 g of KO 2 and 47.0 g of H 2 O, how many grams of O 2 can be produced? 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 120.0 g47.0 g ? g Hide one Based on: KO 2 = g O 2 120.0 g KO 2 40.51 Limiting/Excess Reactant Problem with % Yield

23. 22 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) If a reaction vessel contains 120.0 g of KO 2 and 47.0 g of H 2 O, how many grams of O 2 can be produced? 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 120.0 g47.0 g ? g Based on: KO 2 = g O 2 120.0 g KO 2 40.51 Based on: H 2 O = g O 2 Question if only 35.2 g of O 2 were recovered, what was the percent yield? Hide 47.0 g H 2 O 125.3 Limiting/Excess Reactant Problem with % Yield