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# Fun with Solutions - PowerPoint PPT Presentation

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

You have 2 solutions that you are going to mix together. on to something a little more fun.

• 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.

First things first. Will these 2 solutions react together? on to something a little more fun.

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

Molecular Equation on to something a little more fun.

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

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

The volume is easy. need the number of moles of chloride and the total volume of the solution.

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

• V = 100. mL + 150. mL = 250. mL

Now let’s look at chloride need the number of moles of chloride and the total volume of the solution.

• 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…. need the number of moles of chloride and the total volume of the solution.

• 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 need the number of moles of chloride and the total volume of the solution.

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

Of chloride that is.

The rest is easy. need the number of moles of chloride and the total volume of the solution.

• [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! need the number of moles of chloride and the total volume of the solution.

• - Minimum Volume to precipitate – true stoichiometry

Where do I start? need the number of moles of chloride and the total volume of the solution.

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

Write the double displacement reaction. need the number of moles of chloride and the total volume of the solution.

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

Make sure it is balanced. need the number of moles of chloride and the total volume of the solution.

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

What do you know? need the number of moles of chloride and the total volume of the solution.

• 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

Finally equation) to figure out how many moles of MgCl

• 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! equation) to figure out how many moles of MgCl

• p. 307: 4 and 7