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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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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 H2O

  • 26.5 g of CaCl2 in 150. mL H2O

  • You want to find [Cl-] in the solutions after they are mixed.


  • What to do, what to do?


First things first. Will these 2 solutions react together?

  • Using your knowledge of double displacement reactions you can tell that they will not.


Molecular Equation

KCl (aq) + CaCl2 (aq ) → KCl (aq) + CaCl2(aq)

Aha! The products are the same as the reactants so there IS no reaction.


  • 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.


The volume is easy.

  • When you mix 2 solutions the volumes add so….

  • V = 100. mL + 150. mL = 250. mL


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


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


Add them up

In total you have 0.266 mol + 0.478 mol = 0.744 mol

Of chloride that is.


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


But wait! It gets better!

  • - Minimum Volume to precipitate – true stoichiometry


  • 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?


Where do I start?

  • This problem is based on a knowledge of double displacement reactions so that is a good place to start.


Write the double displacement reaction.

  • MgCl2(aq) + AgNO3 (aq) → AgCl (s) + Mg(NO3)2 (aq)


Make sure it is balanced.

  • MgCl2(aq) + 2 AgNO3 (aq) → 2 AgCl (s) + Mg(NO3)2 (aq)


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


  • 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


Finally

  • You can find the volume of the magnesium chloride solution that contains that number of moles.

  • V =

  • Or 36 mL of MgCl2 solution.


Now you try!

  • p. 307: 4 and 7


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


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