1. WE CANNOT SOLVE OUR PROBLEMS WITH THE SAME THINKING WE USED WHEN WE CREATED THEM. –Einstein-

2. LIMITING REAGENT PROBLEM: A strip of Zn metal weighing 2.00 g is placed in an aqueous solution containing 2.50 g silver nitrate causing the following reaction to occur: Zn (s) + 2 AgNO 3 (aq)  2 Ag (s) + Zn(NO 3 ) 2 (aq) a)Which reagent is the limiting reagent? b)B) How many grams of silver will form? From the equation, 1 mole of Zn reacts with 2 moles of AgNO 3. Let’s calculate how many moles of each we have to start. Zn = 2.00/65.4 = 0.0306 mole AgNO 3 = 2.50/(108 + 14 + 48) = 2.50/170 = 0.0147 mole Since you need 2 moles AgNO 3 for every mole of Zn, your limiting reagent is AgNO 3.

3. SO, LET’S CALCULATE THE AMOUNT OF SILVER FORMED: 2.50 g X g Zn (s) + 2 AgNO 3 (aq)  2 Ag (s) + Zn(NO 3 ) 2 (aq) 2(170) g 2(108) To calculate silver formed 2.5/2(170) = X/2(108) and X = (2.5/340) x 216 = 1.59 g

4. The theoretical yield is how much you could produce in a chemical reaction if your limiting reagent were all used up and the reaction went perfectly. Things almost never work perfectly, so we talk about actual yield and the % theoretical yield. Adipic acid H 2 C 6 H 8 O 4 is used to produce nylon. It is made commercially by a controlled reaction between cyclohexane C 6 H 12 and oxygen O 2 : 2 C 6 H 12 + 5 O 2  2 H 2 C 6 H 8 O 4 + 2 H 2 O Assume you carry out the reaction starting with 25.0 g cyclohexane, and that cyclohexane is the limiting reagent. What is the theoretical yield of adipic acid?

5. Molar mass of cyclohexane is: 6 C = 6 x 12.0 = 72.0 12 H = 12 x 1.01 = 12.1 molar mass = 84.1g Molar mass of adipic acid is: 10 H = 10 x 1.01 = 10.1 6 C = 6 x 12.0 = 72.0 4 O = 4 x 14 = 64.0 molar mass = 146 g 25.0 X 2 C 6 H 12 + 5 O 2  2 H 2 C 6 H 8 O 4 + 2 H 2 O 2(84.1) 2(146) Or, 25/2(84.1) = X/2(146) X = (25/84.1) x 146 = 43.4 g

6. 43.4 g is the theoretical yield. If you obtain 33.5 g for the actual yield, what is the percent yield? Percent yield = (33.5/43.4) x 100 = 77.2 %