html5-img
1 / 15

a.k.a ….having twice as much fun by doing twice as much math 

a.k.a ….having twice as much fun by doing twice as much math . H 2. O 2. 8 H : 8 O. Imagine that paperclips are atoms and that we’re starting with an equal number of hydrogen atoms and oxygen atoms. For example:. 2H 2 + O 2 -> 2H 2 O.

almira
Download Presentation

a.k.a ….having twice as much fun by doing twice as much math 

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. a.k.a….having twice as much fun by doing twice as much math 

  2. H2 O2 8 H : 8 O Imagine that paperclips are atoms and that we’re starting with an equal number of hydrogen atoms and oxygen atoms. For example: 2H2 + O2 -> 2H2O Let’s assemble our paperclip atoms into paperclip molecules.

  3. H2 O2 H2O The product molecules are made out of the available reactant molecules.

  4. H2 O2 H2O Continue to convert reactants to products.

  5. H2O Continue to convert reactants to products. H2 O2

  6. H2O Continue to convert reactants to products. H2 O2

  7. H2O H is the limiting reagent 2 The reaction stops after forming 4 water molecules because there aren’t any more hydrogen molecules to react. H2 O2 The reaction was limited by the amount of hydrogen initially present. Therefore,

  8. H2 O2 H2O + 2 H2 + O2 2 H2O Q: How many grams of water will be formed? A: 4 mol x 18.015 g/mol 4 mol H2 4 mol H2O 72 g H2O 4 mol O2 8 mol H2O Step 1: Assume all of the hydrogen reacts and determine the number of moles of water that would be formed. = 4 mol Step 2: Assume all of the oxygen reacts and determine the number of moles of water that would be formed. = 8 mol Step 3: Multiply the smaller molar amount by the gfm of water to find the number of grams of water that will be formed.

  9. For the reaction below, how many moles of water will be formed if 1.55 moles of hydrogen gas is allowed to react with 0.30 moles of oxygen gas? 2 H2 + O2 2 H2O 1.55 mol H2 1.55 mol H2O 0.30 mol O2 0.60 mol H2O Which reagent is in excess? What is the limiting reagent? O2 H2 How many grams of H2O (gfm = 18.015 g/mol) will be formed from this reaction? 10.8 g of H2O will form 0.60 mol H2O x 18.015 g/mol =

  10. O 2 For the reaction below, how many grams of water will be formed if 1.17 g of hydrogen gas is allowed to react with 7.35 g of oxygen gas? 2 H2 + O2 2 H2O 8.29 g g: 1.17 g 7.35 g gfm: 2.016 g/mol 31.999 g/mol x 18.015 g/mol 0.58 mol 0.58 mol mol: —— 0.23 mol 0.46 mol What is the limiting reagent? How many grams of water will be formed? 8.29 g

  11. And now for something slightly different…

  12. + + +

  13. The Great Candy Caper! Mission: To separate the bag of candy into three identical piles and thereby determine which type of candy is the limiting reactant. Caution: If you eat any of the candy before telling me what the limiting reactant is, you will self-destruct.

  14. The Great Candy Caper! Mission: To separate the bag of candy into four identical piles and thereby determine which type of candy is the limiting reactant. Caution: If you eat any of the candy before telling me what the limiting reactant is, you will self-destruct.

  15. How many grams of the underlined product will be formed? 4 Al + 3O22Al2O3 3H2 + N2 2NH3 CH4 + 2O2CO2 + 2H2O CH2O + O2 CO2 + H2O 2 Al + 3 PbCl2 2AlCl3 + 3Pb Zn + 2HCl  ZnCl2 + H2 2.16 g Al 0.81 g H2 4.53 g NH3 11.20 g N2 1.28 g O2 2.72 g Al2O3 6.42 g CH4 18.02 g CH2O 4.40 g CO2 6.40 g O2 9.60 g O2 5.40 g H2O 5.23 g Zn 1.08 g Al 11.12 g PbCl2 3.55 g AlCl3 2.92 g HCl 0.081 g H2

More Related