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Warmup

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. 3. 2. 2. 3.4 moles Al 2 O 3.

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Warmup

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  1. Warmup 1. Calculate the molar mass of SbI3 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 + O2 Al2O3 6.8 moles Al moles Al2O3 = moles Al 4 3 2 2 3.4 moles Al2O3 4

  2. Rock 'N' Mole "To mole or not to mole, this is the question.“ Molybdenum 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 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 Limiting Reactants Notes Grams given Moles given Moles unknown Grams unknown Molar mass step Mole-mole conversion step Molar mass step

  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. 2 4 3 Ex 1: How many grams of Al2O3 can be formed from 5.81 grams of Al? Al + O2 Al2O3 1. Write/Balance Equation: 2. Figure out Given: 5.81 grams Al Figure out Unknown: ? grams Al2O3 3. Choose correct pathway: g Al  mol Al  mol Al2O3  g Al2O3 4. Numbers! ( mol Al) 1 ( mol Al2O3 ) 2 ( g Al2O3 ) 101.96 5.81 g Al ( g Al) 26.98 ( mol Al) 4 ( mol Al2O3 ) 1 = 11.0 grams Al2O3

  6. 1 1 2 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? H2 + Cl2 HCl 1. Write/Balance Equation: 2. Figure out Given: 3.75 grams Cl2 Figure out Unknown: ? grams HCl 3. Choose correct pathway: g Cl2 mol Cl2  mol HCl g HCl 4. Numbers! ( g HCl) 3.75 g Cl2 ( mol Cl2 ) 1 ( mol HCl ) 2 36.46 1 70.9 1 ( mol HCl ) ( g Cl2 ) ( mol Cl2 ) = 3.86 grams HCl

  7. 2Sb(s) + 3I2(s) → 2SbI3(s Ex 3. What mass of antimony(III) iodide can form from 1.20g Sb ? 2 1.20 g Sb ( mol Sb ) 1 ( mol SbI3) ( g SbI3 ) 502.46 1 121.76 2 ( mol SbI3 ) ( g Sb ) ( mol Sb) = 4.94 g SbI3 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) + 3I2(s) → 2SbI3(s) Ex 4. Determine the theoretical yield (mass in grams) of antimony(III) iodide formed when 1.20g Sb and 2.40 g I2 are mixed. 2.40 g I2 ( mol I2 ) 1 ( mol SbI3) 2 ( g SbI3 ) 502.46 ( g I2 ) 253.8 ( mol I2) 3 ( mol SbI3 ) 1 = 3.17 g SbI3 wait but 1.20 g Sb predicts 4.94 g SbI3 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. 3.17 g SbI3.I2 is the limiting reactant (even though there is a greater amount of it) and is completely used up. Some Sb will be leftover.

  9. 4 3 2 6 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 36.0 g NH3 ( mol NH3 ) 1 ( mol N2) 2 ( g N2 ) 28.02 ( g NH3) 17.04 ( mol NH3) 4 ( mol N2 ) 1 = 29.6 g N2 2 1 28.02 50.0 g O2 ( mol O2 ) ( mol N2) ( g N2 ) 32.00 ( mol N2 ) 1 ( g O2 ) ( mol O2) 3 = 29.2 g N2

  10. 4 3 2 6 29.2 g N2 were produced. Oxygen gas = limiting reactant. Rephrase this question: how much ammonia is required to produce 29.2 g N2 ? Ex 6. From the previous problem, what mass of excess reactant remains? NH3 + O2  N2 + H2O 29.2 g N2 ( mol N2 ) 1 ( mol NH3) 4 ( g NH3 ) 17.04 ( mol NH3 ) ( g N2 ) ( mol N2) 2 1 28.02 = 35.5 g NH3 35.5 g were actually required but 36.0 grams NH3 were used in the reaction. Therefore, the excess NH3 = 0.5 g

  11. Ex 7. In the previous problem, we calculated that 29.2 g N2 should be produced. If 26.7 grams N2 are actually produced, calculate the percent yield of the experiment • percent yield = actual yield x 100 • theoretical yield • percent yield = 26.7 g x 100 • 29.2 g 91.4 %

  12. Conversion flow chart #molecules /atoms /ions given # unknown molecules / atoms/ions Grams given Moles given Moles unknown Grams unknown Molar mass step Mole-mole conversion step Molar mass step

  13. 1 1 2 Ex 8: 6.7 x 1017 molecules chlorine gas can form what mass of hydrochloric acid? Assume an excess of hydrogen gas. H2 + Cl2 HCl Pathway: molec. Cl2 mol Cl2  mol HCl g HCl 6.7 x 1017 molec Cl2 1 ( mol HCl ) 2 36.46 ( mol Cl2 ) ( g HCl) 6.02x 1023 1 ( mol HCl ) 1 ( molec. Cl2 ) ( mol Cl2 ) = 8.1 x 10-5 grams HCl

  14. 1 1 2 Ex 9: 98.7 g Cl2 will completely react with how many moles of hydrogen gas? H2 + Cl2 HCl Pathway: g Cl2 mol Cl2  mol H2 98.7 g Cl2 1 ( mol H2 ) 1 ( mol Cl2 ) 70.9 1 ( g Cl2 ) ( mol Cl2 ) = 1.39 moles H2

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