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Fun with Solutions

Fun with Solutions. Now that you have finished your quiz you are ready to move on to something a little more fun. What could be more fun that doing stoichiometry? No , seriously. You have 2 solutions that you are going to mix together. 19.8 g of KCl in 100. mL H 2 O

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Fun with Solutions

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  1. Fun with Solutions

  2. Now that you have finished your quiz you are ready to move on to something a little more fun. • What could be more fun that doing stoichiometry? No , seriously.

  3. You have 2 solutions that you are going to mix together. • 19.8 g of KCl in 100. mL H2O • 26.5 g of CaCl2 in 150. mL H2O • You want to find [Cl-] in the solutions after they are mixed.

  4. What to do, what to do?

  5. First things first. Will these 2 solutions react together? • Using your knowledge of double displacement reactions you can tell that they will not.

  6. Molecular Equation KCl (aq) + CaCl2 (aq ) → KCl (aq) + CaCl2(aq) Aha! The products are the same as the reactants so there IS no reaction.

  7. If you want to find the concentration of chloride ions you need the number of moles of chloride and the total volume of the solution.

  8. The volume is easy. • When you mix 2 solutions the volumes add so…. • V = 100. mL + 150. mL = 250. mL

  9. Now let’s look at chloride • Now there are 2 sources of chloride ion so you need to find the amount of chloride that each source is going to give you. • Mol Cl- in KCl = 19.8 g KCl x 1 mol KCl x 1 mol Cl-= 0.266 mol • 74.55 g KCl 1 mol KCl

  10. Chloride cont’d…. • And from the second source mol Cl- in CaCl2 = 26.5 g CaCl2 x 1 mol CaCl2 x 2 mol Cl- = 0.478 mol 110.98 g CaCl2 1 mol CaCl2

  11. Add them up In total you have 0.266 mol + 0.478 mol = 0.744 mol Of chloride that is.

  12. The rest is easy. • [Cl-] = mol/V = 0.478 mol/ 0.250 L = 2.98 mol/L • p. 302: 11-14 • For these text questions you need to remember that moles can also be found from solutions using mol = M x V

  13. But wait! It gets better! • - Minimum Volume to precipitate – true stoichiometry

  14. What is the minimum volume of 0.25 M MgCl2 that is required to precipitate all the silver ions in 60. mL of 0.30 M AgNO3?

  15. Where do I start? • This problem is based on a knowledge of double displacement reactions so that is a good place to start.

  16. Write the double displacement reaction. • MgCl2(aq) + AgNO3 (aq) → AgCl (s) + Mg(NO3)2 (aq)

  17. Make sure it is balanced. • MgCl2(aq) + 2 AgNO3 (aq) → 2 AgCl (s) + Mg(NO3)2 (aq)

  18. What do you know? • You know the volume and concentration of the silver containing compound AgNO3. From this you can calculate the number of moles. • x mol AgNO3 = M x V = 0.30 mol/L x 0.060 L = 0.018 mol

  19. Now you use stoichiometry (and your balanced chemical equation) to figure out how many moles of MgCl2 you need to react with 0.018 mol AgNO3. • x mol MgCl2 = = 0.0090 mol MgCl2

  20. Finally • You can find the volume of the magnesium chloride solution that contains that number of moles. • V = • Or 36 mL of MgCl2 solution.

  21. Now you try! • p. 307: 4 and 7

  22. Mrs. Reid should go away more often and leave us fun work like this to do.

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