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 H 2 O

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Fun with Solutions

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Fun with solutions

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

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.


Fun with solutions

  • What to do, what to do?


First things first will these 2 solutions react together

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

Molecular Equation

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

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


Fun with solutions

  • 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

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 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

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

Add them up

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

Of chloride that is.


The rest is easy

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

But wait! It gets better!

  • - Minimum Volume to precipitate – true stoichiometry


Fun with solutions

  • 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

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

Write the double displacement reaction.

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


Make sure it is balanced

Make sure it is balanced.

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


What do you know

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


Fun with solutions

  • 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

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

Now you try!

  • p. 307: 4 and 7


Fun with solutions

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


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