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Limiting Reagents

Limiting Reagents. Note: it would be greatly to your benefit to have gone through my Stoichiometry tutorial before attempting this one (bfrench.ppt). Bobby French pd4. Easy. Not so easy.

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Limiting Reagents

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  1. Limiting Reagents Note: it would be greatly to your benefit to have gone through my Stoichiometry tutorial before attempting this one (bfrench.ppt) Bobby French pd4 Easy Not so easy

  2. You need help with limiting reagents, or you wouldn’t be here. Try this tutorial; it might help! Ok, the problem is: Ca + HCl CaCl2 + H2 Your job: find the limiting reagent and how much calcium chloride will be made from 80.2g Ca and 73.0g HCl What’s the first step? Start with 80.2g Ca and go from there Balance the equation Start with 73.0g HCl and go from there

  3. Correct! Ok, the problem is: Ca + 2HCl CaCl2 + H2 Your job: find the limiting reagent and how much calcium chloride will be made from 80.2g Ca and 73.0g HCl What’s the next step? Go from 80.2g Ca to g of HCl and see if that number Is larger Find out how much product will be made from 80.2g Ca or from 73.0g HCl Convert both to moles of H2

  4. Correct! Ok, the problem is: Ca + 2HCl CaCl2 + H2 Your job: find the limiting reagent and how much calcium chloride will be made from 80.2g Ca and 73.0g HCl How many moles of CaCl2 will 80.2g Ca make? 2.00 1.53 1.0

  5. Correct! Ok, the problem is: Ca + 2HCl CaCl2 + H2 Your job: find the limiting reagent and how much calcium chloride will be made from 80.2g Ca and 73.0g HCl How many moles of CaCl2 will 73.0g HCl make? 2.0 3.00 1.00

  6. Correct! Ok, the problem is: Ca + 2HCl CaCl2 + H2 Your job: find the limiting reagent and how much calcium chloride will be made from 80.2g Ca and 73.0g HCl So now that we know that, which one is the limiter? HCl H2 Ca

  7. Correct! Ok, the problem is: Ca + 2HCl CaCl2 + H2 Your job: find the limiting reagent and how much calcium chloride will be made from 80.2g Ca and 73.0g HCl So then, how many moles of product are produced? 1.00 2.00 1.530

  8. Correct! Ok, the problem is: Ca + 2HCl CaCl2 + H2 Your job: find the limiting reagent and how much calcium chloride will be made from 80.2g Ca and 73.0g HCl So then, how many grams of product are produced? 110 111.1 111

  9. Correct! Ok, the problem is: Ca + 2HCl CaCl2 + H2 Your job: find the limiting reagent and how much calcium chloride will be made from 80.2g Ca and 73.0g HCl Summarize what the answers are. HCl, 1.53 mol, 111g HCl, 1.00mol, 111g HCl, 1.00mol, 110g

  10. Good job! You finished the easy limiting reagent tutorial! You should try the other one now. Not so easy? End show?

  11. Good! If you know that already, don’t bother with the easy stuff. Good job! Back to the easy anyways? Not so easy one now?

  12. You’ve been Lego-ified! Go back to your work and get it right!

  13. Confident, are you? Good. The next problem: Al(OH)3 + HCl AlCl3 + H2O Find out which reactant is the limiting reagent and find how much AlCl3 is produced. You will also find out how much of the excess reagent is left over. You will do this by going from one reactant to the other. You are given 117.0g Al(OH)3 and 182.5g HCl. What is your first move? Go from either reactant to AlCl3 Use Stoichiometry to go from Al(OH)3 to HCl Balance the chemical equation

  14. Correct! The problem: Al(OH)3 + 3HCl AlCl3 + 3H2O Find out which reactant is the limiting reagent, how much excess reagent is left over, and find out how much AlCl3 is produced. You will do this by going from one reactant to the other. You are given 117.0g Al(OH)3 and 182.5g HCl. What’s your next move? Go from Al(OH)3 to HCl Go from either reactant to AlCl3 Re-balance the chemical equation

  15. Correct! The problem: Al(OH)3 + 3HCl AlCl3 + 3H2O Find out which reactant is the limiting reagent, how much excess reagent is left over, and find out how much AlCl3 is produced. You will do this by going from one reactant to the other. You are given 117.0g Al(OH)3 and 182.5g HCl. You get 164.3g HCl when you go one way, so what now? Try the other way Say Al(OH)3 is the excess reagent Say HCl is the excess reagent

  16. Excellent! That is correct, but I have to go slowly for some people. Since you are smarter than that, you get to go ahead. Yay! Go on!

  17. Correct! The problem: Al(OH)3 + 3HCl AlCl3 + 3H2O Find out which reactant is the limiting reagent, how much excess reagent is left over, and find out how much AlCl3 is produced. You will do this by going from one reactant to the other. You are given 117.0g Al(OH)3 and 182.5g HCl. You get 164.3g HCl when you go one way, and 130.0g Al(OH)3 the other way, so what now? Say that neither one is the excess reagent Say Al(OH)3 is the excess reagent Say HCl is the excess reagent

  18. Correct! The problem: Al(OH)3 + 3HCl AlCl3 + 3H2O Find out which reactant is the limiting reagent, how much excess reagent is left over, and find out how much AlCl3 is produced. You will do this by going from one reactant to the other. You are given 117.0g Al(OH)3 and 182.5g HCl. You get 164.3g HCl when you go one way, and 130.0g Al(OH)3 the other way, so you know that HCl is the excess reagent. How much of it is left over? 0.5000mol 1.500mol 0.500mol Explanation:

  19. Correct! The problem: Al(OH)3 + 3HCl AlCl3 + 3H2O Find out which reactant is the limiting reagent, how much excess reagent is left over, and find out how much AlCl3 is produced. You will do this by going from one reactant to the other. You are given 117.0g Al(OH)3 and 182.5g HCl. You get 164.3g HCl when you go one way, and 130.0g Al(OH)3 the other way, so you know that HCl is the excess reagent. You have 0.5000mol HCl left over; how much AlCl3 is produced? 0.5000mol 1.5000mol 1.500mol

  20. Correct! The problem: Al(OH)3 + 3HCl AlCl3 + 3H2O Find out which reactant is the limiting reagent, how much excess reagent is left over, and find out how much AlCl3 is produced. You will do this by going from one reactant to the other. You are given 117.0g Al(OH)3 and 182.5g HCl. You get 164.3g HCl when you go one way, and 130.0g Al(OH)3 the other way. Summarize the answers: HCl, 1.500mol HCl, and 1.500mol AlCl3 Al(OH)3, 0.5000mol Al(OH)3, and 0.5000mol AlCl3 HCl, 0.5000 mol HCl, and 1.500mol AlCl3

  21. Excellent! You finished the “Not so easy” tutorial! You get no prize. Still, you might have learned something, and that’s worth it. Good for you. End show

  22. You thought that you could get through this one that easily, eh? Go back to the easy one! (You got Lego-ified again!) Fine…. back to your mistake Yes, get me to easy!

  23. Ok, you want an explanation. We started with 117.0g Al(OH)3, and 182.5g HCl, and when we went from Al(OH)3 to HCl, we got 164.3g HCl. When we went from HCl to Al(OH)3, we got 130.0g Al(OH)3. So, that means that if we just had 182.5g HCl, we would need 130.0g Al(OH)3 to react fully with the HCl. Since that is more than we started with, we cannot have a reaction that uses up all of the HCl with 182.5g HCl and 117.0g Al(OH)3. Also, since we got 164.3g HCl when we started with Al(OH)3, that means that a “full” reaction (going by Al(OH)3) would only need 164.3g HCl, so we would have some HCl left over. That means that HCl is the limiting reagent. Once you figure that out, it’s just straight stoichiometry from there. Back

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