1. LIMITING REACTANT The reactant that gives the least number of product moles “limits” the reaction. To understand this concept, let’s suppose you were an elf working for Santa Claus and your job was to make candy canes. You take one red stick and one white stick then twist them around to make one candy cane. The ratio of red to white is 1:1. Time is ticking and you find that you have 24 red stick and 17 white sticks left. What is the maximum number of candy canes you can make? *MOLES* The answer is 17 candy canes! The white sticks “limit” the amount of product you could make. In chemistry we do not use sticks but *MOLES* to determine which starting material will limit the maximum amount of product that can be produced in a chemical reaction.

2. LIMITING REACTANT a) Sodium metal reacts with oxygen to produce sodium oxide. If 5.00 g of sodium reacted with 5.00 grams of oxygen, how many grams of product is formed? 4 Na (s) + O 2(g)  2 Na 2 O (s) Start with what is given, calculate the amount of product that can be theoretically made but do it twice (once for each reactant): 6.74 g of Na 2 O 5.00g Na ( 1 mole Na ) ( 2 mole Na 2 O )( 62 g Na 2 O ) = 6.74 g of Na 2 O 23 g Na 4 mole Na 1 mol Na 2 O 19.38 g of Na 2 O 5.00g O 2 ( 1 mole O 2 ) ( 2 mole Na 2 O )( 62 g Na 2 O ) = 19.38 g of Na 2 O 32 g O 2 1 mole O 2 1 mol Na 2 O Notice you can not have two different masses produced for the same product in one reaction vessel! So in this case, Na (sodium) “limits” how much sodium oxide is produced. The correct answer is 6.74 g of sodium oxide. Wrong answer

3. LIMITING REACTANT b) How much oxygen was used in this reaction and how much of each reactant was leftover (in excess)? 4 Na (s) + O 2(g)  2 Na 2 O (s) There are two methods used to answer this question. The Law of Conservation of mass and Stoichiometry. The amount of O 2 used to make 6.74 g of Na 2 O is calculated by: 1.74 g of O 2 was used 5.00g Na ( 1 mole Na ) ( 1 mole O 2 )( 32 g O 2 ) = 1.74 g of O 2 was used 23 g Na 4 mole Na 1 mol O 2 Or use the Law of Conservation of mass: Mass of product (6.74 g) – mass of limiting reactant (5.00 g) = mass of other reactant, in this case oxygen (1.74 g). The amount of oxygen (O 2 ) leftover can be calculated by subtracting the starting mass of oxygen from the used mass. 5.00g – 1.74 g = 3.26 g The amount of sodium (Na) leftover at the end of the reaction is “0.00 g” (zero), since it was the limiting reactant and was completely consumed in the reaction.

4. LIMITING REACTANT How many grams of solid are formed when 10.0 g of lead reacts with 10.0 g of phosphoric acid? 3 Pb+2 H 3 PO 4  Pb 3 (PO 4 ) 2 (s) + 3 H 2 (g) Start with what is given but do it twice: 13.1 g Pb 3 (PO 4 ) 2 10.0g Pb ( 1 mole Pb ) ( 1 mole Pb 3 (PO 4 ) 2 )( 811 g Pb 3 (PO 4 ) 2 ) = 13.1 g Pb 3 (PO 4 ) 2 207 g 3 mole Pb 1 mole Pb 3 (PO 4 ) 2 41.4 g P b 3 (PO 4 ) 2 10.0g H 3 PO 4 ( 1 mole H 3 PO 4 )( 1 mol Pb 3 (PO 4 ) 2 )( 811 g Pb 3 (PO 4 ) 2 )= 41.4 g P b 3 (PO 4 ) 2 98 g 2 mole H 3 PO 4 1 mole Pb 3 (PO 4 ) 2 You can not have two different answers for one question so in this case lead “limits” how much lead(II) phosphate that can be produced. The correct answer is 13.1 g. Wrong answer

5. PRACTICE PROBLEM #19 1. How much AgCl product will be produced if 100.00 g of BaCl 2 reacted with excess AgNO 3 ? 2. How many moles of carbon dioxide could be produced from 220.0 g of C 2 H 2 and 545.0 g of O 2 ? 3. How many grams of CO 2 can be produced by the reaction of 35.5 grams of C 2 H 2 and 45.9 grams of O 2 ? 4. In the reaction between CH 4 and O 2, if 25.0 g of CO 2 are produced, what is the minimum amount of each reactant needed? 5. Cu + 2 AgNO 3  Cu(NO 3 ) 2 + 2 Ag. When 10.0 g of copper was reacted with 60.0 g of silver nitrate solution, 30.0 g of silver was obtained. What is the percent yield of silver obtained? 137.57 g 50.5 g 13.63 mol 9.09 g of CH 4 & 36.4 g of O 2 88.3%

6. GROUP STUDY PROBLEM #19 ______1. Which reactant will produce the least amount of AgCl product if reacted with AgNO 3 ? a)100.00 g BaCl 2 b) 400.0 g NaClc) 200.0 g CsCl ______2. How many moles of CO 2 can be produced by the reaction of 5.0 grams of C 2 H 4 and 12.0 grams of O 2 ? ______3. How many grams of carbon dioxide could be produced from 2.0 g of C 2 H 4 and 5.0 g of O 2 ? ______4. In the reaction between CH 4 and O 2, if 18.0 g of CO 2 are produced, how many grams of water are produced? ______5. Cu + 2 AgNO 3  Cu(NO 3 ) 2 + 2 Ag. When 50.0 g of copper was reacted with 300.0 g of silver nitrate solution, 149 g of silver was obtained. What is the percent yield of silver obtained?