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Stoichiometry: Finding the Limiting Reactant in a Calcium Hydroxide and Hydrochloric Acid Reaction

This guide explains how to determine the limiting reactant and find the mass of water produced in the reaction between calcium hydroxide (Ca(OH)2) and hydrochloric acid (HCl). Beginning with ensuring that the chemical equation is balanced, we step through the calculation of moles and masses involved in the reaction. With detailed steps on setting up the equations and identifying reactants, this resource simplifies the stoichiometric process for students and chemistry enthusiasts alike.

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Stoichiometry: Finding the Limiting Reactant in a Calcium Hydroxide and Hydrochloric Acid Reaction

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Presentation Transcript

  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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