# Fun with Solutions - PowerPoint PPT Presentation

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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

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.