1. 1 Question of the Day A solution containing 3.50 g sodium phosphate is mixed with a solution containing 6.40 g barium nitrate. After the reaction is complete, 4.85 g barium phosphate are collected. 1. What is the limiting reagent? 2. What is the theoretical yield of barium phosphate? 3. What is the percent yield of barium phosphate? 2Na 3 PO 4(aq) + 3Ba(NO 3 ) 2(aq)  6NaNO 3(aq) + Ba 3 (PO 4 ) 2(s) molar masses: Na 3 PO 4 = 163.9 Ba(NO 3 ) 2 = 261.3 Ba 3 (PO 4 ) 2 = 601.9 Day 1 12-21

2. 3.15 Urea [(NH 2 ) 2 CO] is prepared by reacting ammonia with carbon dioxide: In one process, 637.2 g of NH 3 are treated with 1142 g of CO 2. (a) Which of the two reactants is the limiting reagent? (b) Calculate the mass of (NH 2 ) 2 CO formed. (c) How much excess reagent (in grams) is left at the end of the reaction? Question of the Day Day 2 12-22

3. 3.15 (a)Strategy The reactant that produces fewer moles of product is the limiting reagent because it limits the amount of product that can be formed. How do we convert from the amount of reactant to amount of product? Perform this calculation for each reactant, then compare the moles of product, (NH 2 ) 2 CO, formed by the given amounts of NH 3 and CO 2 to determine which reactant is the limiting reagent.

4. 3.15 Solution We carry out two separate calculations. First, starting with 637.2 g of NH 3, we calculate the number of moles of (NH 2 ) 2 CO that could be produced if all the NH 3 reacted according to the following conversions: Combining these conversions in one step, we write

5. 3.15 Second, for 1142 g of CO 2, the conversions are The number of moles of (NH 2 ) 2 CO that could be produced if all the CO 2 reacted is It follows, therefore, that NH 3 must be the limiting reagent because it produces a smaller amount of (NH 2 ) 2 CO.

6. 3.15 (b) Strategy We determined the moles of (NH 2 ) 2 CO produced in part (a), using NH 3 as the limiting reagent. How do we convert from moles to grams? Solution The molar mass of (NH 2 ) 2 CO is 60.06 g. We use this as a conversion factor to convert from moles of (NH 2 ) 2 CO to grams of (NH 2 ) 2 CO: Check Does your answer seem reasonable? 18.71 moles of product are formed. What is the mass of 1 mole of (NH 2 ) 2 CO?

7. 3.15 (c) Strategy Working backward, we can determine the amount of CO 2 that reacted to produce 18.71 moles of (NH 2 ) 2 CO. The amount of CO 2 left over is the difference between the initial amount and the amount reacted. Solution Starting with 18.71 moles of (NH 2 ) 2 CO, we can determine the mass of CO 2 that reacted using the mole ratio from the balanced equation and the molar mass of CO 2. The conversion steps are

8. 3.15 Combining these conversions in one step, we write The amount of CO 2 remaining (in excess) is the difference between the initial amount (1142 g) and the amount reacted (823.4 g): mass of CO 2 remaining = 1142 g − 823.4 g = 319 g

9. Day 4 1-4 1. Which reactant determines your theoretical yield?

10. Day 4 1-4

11. Example 3.16 The reaction between alcohols and halogen compounds to form ethers is important in organic chemistry, as illustrated here for the reaction between methanol (CH 3 OH) and methyl bromide (CH 3 Br) to form dimethylether (CH 3 OCH 3 ), which is a useful precursor to other organic compounds and an aerosol propellant. This reaction is carried out in a dry (water-free) organic solvent, and the butyl lithium (LiC 4 H 9 ) serves to remove a hydrogen ion from CH 3 OH. Butyl lithium will also react with any residual water in the solvent, so the reaction is typically carried out with 2.5 molar equivalents of that reagent. How many grams of CH 3 Br and LiC 4 H 9 will be needed to carry out the preceding reaction with 10.0 g of CH 3 OH?

12. Example 3.16 Solution We start with the knowledge that CH 3 OH and CH 3 Br are present in stoichiometric amounts and that LiC 4 H 9 is the excess reagent. To calculate the quantities of CH 3 Br and LiC 4 H 9 needed, we proceed as shown in Example 3.14.

13. Period 3 Day 5 1-5 1. Balance the following equation: Al + Cr 2 O 3  Al 2 O 3 + Cr

14. Day 5 1-5 Test will be Thursday and Friday of this week and will cover ALL of Chapter 3!!!

15. Example 3.17 Titanium is a strong, lightweight, corrosion-resistant metal that is used in rockets, aircraft, jet engines, and bicycle frames. It is prepared by the reaction of titanium(IV) chloride with molten magnesium between 950°C and 1150°C: In a certain industrial operation 3.54 × 10 7 g of TiCl 4 are reacted with 1.13 × 10 7 g of Mg. (a)Calculate the theoretical yield of Ti in grams. (b)Calculate the percent yield if 7.91 × 10 6 g of Ti are actually obtained.

16. Example 3.17 (a)Strategy Because there are two reactants, this is likely to be a limiting reagent problem. The reactant that produces fewer moles of product is the limiting reagent. How do we convert from amount of reactant to amount of product? Perform this calculation for each reactant, then compare the moles of product, Ti, formed.

17. Example 3.17 Solution Carry out two separate calculations to see which of the two reactants is the limiting reagent. First, starting with 3.54 × 10 7 g of TiCl 4, calculate the number of moles of Ti that could be produced if all the TiCl 4 reacted. The conversions are so that

18. Example 3.17 Next, we calculate the number of moles of Ti formed from 1.13 × 10 7 g of Mg. The conversion steps are And we write Therefore, TiCl 4 is the limiting reagent because it produces a smaller amount of Ti.

19. Example 3.17 The mass of Ti formed is (b) Strategy The mass of Ti determined in part (a) is the theoretical yield. The amount given in part (b) is the actual yield of the reaction.

20. Example 3.17 Solution The percent yield is given by Check Should the percent yield be less than 100 percent?

21. Period 1 Day 5 1-5 1. Balance the following equation: Al + Cr 2 O 3  Al 2 O 3 + Cr

22. Day 5 1-5 Test will be Thursday and Friday of this week and will cover ALL of Chapter 3!!!

23. http://wps.prenhall.com/esm_brow n_chemistry_9/2/662/169519.cw/ind ex.htmlhttp://wps.prenhall.com/esm_brow n_chemistry_9/2/662/169519.cw/ind ex.html - Chapter 3 Problem Solving # 1

24. Period 8 Day 5 1-5 1. Balance the following equation: Al + Cr 2 O 3  Al 2 O 3 + Cr

25. Day 5 1-5 Test will be Thursday and Friday of this week and will cover ALL of Chapter 3!!!

26. 26 Question of the Day A solution containing 3.50 g sodium phosphate is mixed with a solution containing 6.40 g barium nitrate. After the reaction is complete, 4.85 g barium phosphate are collected. 1. What is the limiting reagent? 2. What is the theoretical yield of barium phosphate? 3. What is the percent yield of barium phosphate? 2Na 3 PO 4(aq) + 3Ba(NO 3 ) 2(aq)  6NaNO 3(aq) + Ba 3 (PO 4 ) 2(s) Day 1 12-21 Names ___________________________________________ ___________________________________________

27. 27 Question of the Day A solution containing 3.50 g sodium phosphate is mixed with a solution containing 6.40 g barium nitrate. After the reaction is complete, 4.85 g barium phosphate are collected. 1. What is the limiting reagent? 2. What is the theoretical yield of barium phosphate? 3. What is the percent yield of barium phosphate? 2Na 3 PO 4(aq) + 3Ba(NO 3 ) 2(aq)  6NaNO 3(aq) + Ba 3 (PO 4 ) 2(s) Day 1 12-21 Names ___________________________________________ ___________________________________________