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Stoichiometry

Stoichiometry. Finding the Limiting Reactant. THE PROBLEM. What mass of water is formed when 40.35 g of Ca(OH) 2 reacts with 20.0 g of HCl, with a byproduct of CaCl 2 ?. 1 st step:. MAKE SURE YOUR EQUATION IS BALANCED. Ca(OH) 2 + HCl CaCl 2 + H 2 O.

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Stoichiometry

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  1. Stoichiometry Finding the Limiting Reactant

  2. THE PROBLEM What mass of water is formed when 40.35 g of Ca(OH)2 reacts with 20.0 g of HCl, with a byproduct of CaCl2 ?

  3. 1st step: MAKE SURE YOUR EQUATION IS BALANCED Ca(OH)2 + HCl CaCl2 + H2O a. 2Ca(OH)2 + HCl 2CaCl2 + H2O b. Ca(OH)2 + 2HCl 2CaCl2 + H2O c. Ca(OH)2 + 2HCl CaCl2 + 2H2O

  4. Wrong! Back to the problem

  5. 2nd step:SET UP THE EQUATION • what should be the first part of your first equation? a. 40.35 g Ca(OH)2 b. 88.08 g Ca(OH) b. 88.08 g Ca(OH)2 c. 40.35 mol Ca(OH) c. 40.35 mol Ca(OH)2

  6. Wrong! Back to the problem

  7. 3rd step:MORE OF THE EQUATION • What is the next part of the equation? a. b. c. 36.46 g Ca(OH)2 1 mol Ca(OH)2 1 mol Ca(OH)2 88.08 g Ca)OH)2 1 mol Ca(OH)2 40.35 g Ca(OH)2

  8. Wrong! Back to the problem

  9. 4th step:SOME MORE EQUATION What is the next part of the equation? a. b. c. 2 mol H2O 1 mol Ca(OH)2 1 mol Ca(OH)2 2 mol H2O 1 mol H2O 2 mol Ca(OH)2

  10. Wrong! Back to problem

  11. 5th step: and even more equation What is the next part in the equation? c. a. b. 18.02 g H2O 1 mol H20 18.02 mol H2O 1 mol H20 1 mol H2O 18.02 g H2O

  12. Wrong! Back to problem

  13. 6th step: THE EQUATION Your problem should now look like this: 40.35 g Ca(OH)2 x 1 mol Ca(OH)2 x 2 mol H2O x 18.02 g H2O = 88.08 g Ca(OH)2 1 mol Ca(OH)2 1 mol H2O What do you get as your answer? c. a. b. 12.65 g H2O 16.51 mol H2O 16.51 g H2O

  14. Wrong! Back to problem

  15. Great Work! Now work on the second equation to find the limiting reactant The problem: What mass of water is formed when 40.35 g of Ca(OH)2 reacts with 20.0 g of HCl, with a byproduct of CaCl2 ? 1st step: What is the first part of the second equation? a.88.08 g HCl b. 40.35 g Ca(OH)2 c. 20.0 g HCl

  16. Wrong! Back to problem

  17. 2nd step: What is the next part of the equation? a. b. c. 1 mol HCl 36.46 g HCl 36.46 g HCl 1 mol HCl 36.46 g Ca(OH)2 1 mol Ca(OH)2

  18. Wrong! Back to problem

  19. 3rd step: What is the next part of the equation? a. b. c. 2 mol H2O 2 mol HCl 2 mol HCl 2 mol H2O 2 mol H2O 1 mol HCl

  20. Wrong! Back to problem

  21. 4th step: What is the next part of the equation? a. b. c. 1 mol H2O 18.02 g H2O 18.02 g H2O 1 mol H2O 36.46 g HCl 1 mol H2O

  22. Wrong! Back to problem

  23. 5th step: This is the correct equation: 1 mol HCl 36.46 g HCl 2 mol H2O 2 mol HCl 18.02 g H2O 1 mol H2O 20.0 g HCl x x x = What is the answer to this problem? a. b. c. 16.51 g H2O 18.02 g HCl 9.88 g H2O

  24. Wrong! Back to problem

  25. Final Step: What is the limiting reactant? a. H2O b. HCl c. Ca(OH)2

  26. Wrong! Back to problem

  27. Yay! You have found the limiting reactant!!

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