1. Warmup 1. Calculate the molar mass of SbI 3 502.46 g/mole 2. According to the equation below, how many moles of aluminum oxide can be made from 6.8 moles aluminum? Al + O 2  Al 2 O 3 6.8 moles Al moles Al 2 O 3 = moles Al 4 32 4 2 3.4 moles Al 2 O 3

2. a. What is Avogadro's favorite kind of music? b. What line from Shakespeare do high school moles have to memorize? c. What element do moles love to study in chemistry? remolte control remoletly sophmole Rock 'N' Mole "To mole or not to mole, this is the question.“ Molybdenum A device used by moles to watch Bill Nye the Science Guy and the Discovery Channel. Obscurely having to do with a mole. Anyone in the tenth grade who is taking Chemistry already!

3. Conversion flow chart Grams given Moles given Moles unknown Grams unknown Molar mass step Molar mass step Mole-mole conversion step Limiting Reactants Notes

4. Problem: If you start with X grams of Givenium, how many grams of Unknownium will be made? X grams of given (1 mole given )( B moles unknown)(D grams of unknown) = (A grams of given) ( C moles given ) ( 1 mole unknown ) Key: Given = the compound whose amount is given to you in the problem Unknown = compound whose amount you are looking for X = grams that the problem gives you A = molar mass of the compound given in the problem (see periodic table) B and C = numbers from molar ratio, found in the chemical equation D = molar mass of the unknown compound from the problem (see periodic table)

5. Ex 1: How many grams of Al 2 O 3 can be formed from 5.81 grams of Al? Al + O 2  Al 2 O 3 1. Write/Balance Equation: 2. Figure out Given: 5.81 grams Al Figure out Unknown: ? grams Al 2 O 3 3. Choose correct pathway:g Al  mol Al  mol Al 2 O 3  g Al 2 O 3 4. Numbers! 5.81 g Al ( g Al 2 O 3 )( mol Al)( mol Al 2 O 3 ) ( g Al)( mol Al)( mol Al 2 O 3 ) 1 26.98 2 41 101.96 = 11.0 grams Al 2 O 3 43 2

6. Ex 2: Hydrogen gas and chlorine gas react to form hydrochloric acid ( HCl ). How many grams of the product will form in the reaction if 3.75 grams of chlorine gas is used ? H 2 + Cl 2  HCl 2. Figure out Given: 3.75 grams Cl 2 Figure out Unknown: ? grams HCl 3. Choose correct pathway:g Cl 2  mol Cl 2  mol HCl  g HCl 4. Numbers! 3.75 g Cl 2 ( mol Cl 2 )( mol HCl ) ( g Cl 2 )( mol Cl 2 ) ( mol HCl ) 1 70.9 2 1 1 36.46 = 3.86 grams HCl ( g HCl) 1. Write/Balance Equation: 112

7. 2Sb(s) + 3I 2 (s) → 2SbI 3 (s Ex 3. What mass of antimony(III) iodide can form from 1.20g Sb ? 1.20 g Sb( mol Sb )( mol SbI 3 ) ( g Sb )( mol Sb) ( mol SbI 3 ) ( g SbI 3 ) 502.46 121.76 1 1 2 2 = 4.94 g SbI 3 Wait! We know how much Sb we have. But what if there is just a tiny, tiny amount of iodine available? Hmmmm… Limiting Reactant – reactant that is used up first in a reaction; it limits how much product you can make!!!

8. 2Sb(s) + 3I 2 (s) → 2SbI 3 (s) Ex 4. Determine the theoretical yield (mass in grams) of antimony(III) iodide formed when 1.20g Sb and 2.40 g I 2 are mixed. 5. Do another calculation with the other amount given 6. Choose the lower amount…the higher amount is not possible. The theoretical yield is calculated based upon the limiting reactant. 2.40 g I 2 ( mol I 2 )( mol SbI 3 ) ( g I 2 )( mol I 2 )( mol SbI 3 ) ( g SbI 3 ) 502.46 1253.8 12 3 = 3.17 g SbI 3 wait but 1.20 g Sb predicts 4.94 g SbI 3 3.17 g SbI 3. I 2 is the limiting reactant (even though there is a greater amount of it) and is completely used up. Some Sb will be leftover.

9. Ex 5. Suppose 36.0 grams of ammonia and 50.0 grams of oxygen react to form nitrogen gas and water. Calculate the mass of nitrogen gas that could be formed. NH3 + O2  N2 + H2O 3462 36.0 g NH 3 ( mol NH 3 )( mol N 2 ) ( g NH 3 )( mol NH 3 ) ( g N 2 ) 50.0 g O 2 ( mol O 2 )( mol N 2 ) ( g O 2 )( mol O 2 ) ( g N 2 ) 2 4 2 3 ( mol N 2 ) 17.04 1 32.00 1 28.02 1 1 = 29.2 g N 2 = 29.6 g N 2

10. percent yield = actual yield x 100 theoretical yield Ex 6. In the previous problem, we calculated that 29.2 g N 2 should be produced. If 26.7 grams N 2 are actually produced, calculate the percent yield of the experiment percent yield = 26.7 g x 100 29.2 g 91.4 %