2. 1 STOICHIOMETRY TUTORIAL Paul Gilletti

3. 2 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.

4. 3 (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.

5. 4 Warm-Up: 3AgNO 3 + Na 3 PO 4  Ag 3 PO 4 + 3NaNO 3 If 200. g of silver nitrate are used, how many grams of silver phosphate are produced? 5:004:594:584:574:564:554:544:534:524:514:504:494:484:474:464:454:444:434:424:414:404:394:384:374:364:354:344:334:324:314:304:294:284:274:264:254:244:234:224:214:204:194:184:174:164:154:144:134:124:114:104:094:084:074:064:054:044:034:024:014:003:593:583:573:563:553:543:533:523:513:503:493:483:473:463:453:443:433:423:413:403:393:383:373:363:353:343:333:323:313:303:293:283:273:263:253:243:233:223:213:203:193:183:173:163:153:143:133:123:113:103:093:083:073:063:053:043:033:023:013:002:592:582:572:562:552:542:532:522:512:502:492:482:472:462:452:442:432:422:412:402:392:382:372:362:352:342:332:322:312:302:292:282:272:262:252:242:232:222:212:202:192:182:172:162:152:142:132:122:112:102:092:082:072:062:052:042:032:022:012:001:591:581:571:561:551:541:531:521:511:501:491:481:471:461:451:441:431:421:411:401:391:381:371:361:351:341:331:321:311:301:291:281:271:261:251:241:231:221:211:201:191:181:171:161:151:141:131:121:111:101:091:081:071:061:051:041:031:021:011:00 YOU HAVE ONLY ONE MINUTE LEFT ON THIS QUESTION 0:590:580:570:560:550:540:530:520:510:500:490:480:470:460:450:440:430:420:410:400:390:380:370:360:350:340:330:320:310:300:290:280:270:260:250:240:230:220:210:200:190:180:170:160:150:140:130:120:110:100:090:080:070:060:050:040:030:020:010:00 Objective: SWBAT use stoichiometry to determine limiting reactants and theoretical yield. 200. g AgNO 3 X 1 mol AgNO 3 169.9 g AgNO 3 X 1 mol Ag 3 PO 4 3 mol AgNO 3 X 418.7 g Ag 3 PO 4 1 mol Ag 3 PO 4 =164.3 g Ag 3 PO 4

6. 5 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.

7. 6 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

8. 7 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

9. 8 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 :

10. 9 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!

11. 10 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

12. 11 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

13. 12 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

14. 13 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 47.0 g H 2 O 125.3 Determine how many grams of Water were left over. The Difference between the above amounts is directly RELATED to the XS H 2 O. 125.3 - 40.51 = 84.79 g of O 2 that could have been formed from the XS water. = g XS H 2 O 84.79 g O 2 31.83

15. 14 Practice Problems-Green Book P. 119-120-Limiting reactants problems # 1-5 Answers are in the back of the book (p. 326) if you would like to check your work. Please SHOW ALL WORK FOR FULL CREDIT. This includes units.

16. 15 11-2 Practice Problems Each person will have one problem. Solve each problem on the blue sheet of paper with the number of the problem. When you have finished, check your answer with Mrs. Basile and then put your paper in the lab area, number side up. Try some of the other problems, check your answers with the papers in the lab area.

17. 16 Wrap Up Summarize the steps to solving a stoichiometry problem. Why is stoichiometry important?