1. Stoichiometry with Solutions
Combining what you’ve just learned with what you are already good at 

2. Stoichiometry with Solutions
Now that you know about concentration, you have another fact you can use to work with moles.
Because Molarity = moles/Liter, this can be used as a conversion factor.
For example, how many moles of NaCl are in 0.25L of a 2.0M solution? Remember, the Molarity gives you your conversion factor.
Set up the problem using dimensional analysis:
0.25 L x 2.0 mol = 0.50 mol NaCl
1.0 L
Write molarity as a fraction

3. Try another
How many moles of iron (III) chloride are in 0.32 L of a 1.5 M solution?
0.32 L x 1.5 mol = 0.48 mol FeCl3
1.0 L
And one more – how many moles of silver nitrate are in 0.50 L of a M solution?
0.50 L x 0.05 mol = mol AgNO3

4. Once you have determined the number of moles, you can do any other conversion you already know about moles (mass, mole ratio in a reaction, etc).
That means – we have an even-more-improved mole map!

5. Even-more-improved mole map
Molarity written as a fraction
Molarity written as a fraction

6. AgNO3 (aq) + HCl (aq)  AgCl (s) + HNO3 (aq)
Remember – to get from volume of a solution to moles, you need the concentration (the Molarity)!
Consider the following reaction:
AgNO3 (aq) + HCl (aq)  AgCl (s) + HNO3 (aq)
If 0.1 L of a 0.1 M solution of AgNO3 is mixed with unlimited HCl, how many grams of AgCl will be made?
0.1 L x 0.1 mol AgNO3 x 1 mol AgCl x g AgCl = g AgCl
1.0 L mol AgNO mol AgCl
Trace the path taken on the mole map above.

7. Try another one:
If 0.2 L of a 1.0 M solution of HCl react with unlimited silver nitrate, how many grams of AgCl will be made?
0.2 L x 1.0 mol HCl x 1 mol AgCl x g AgCl = g AgCl
1.0 L mol HCl mol AgCl

8. FeCl3 (aq) + 3 NaOH (aq)  Fe(OH)3 (s) + 3 NaCl (aq)
And one more:
A solution of 0.2 M iron (III) chloride reacts with aqueous sodium hydroxide to produce a precipitate of iron (III) hydroxide and aqueous sodium chloride.
Write and balance the equation.
FeCl3 (aq) + 3 NaOH (aq)  Fe(OH)3 (s) + 3 NaCl (aq)

9. FeCl3 (aq) + 3 NaOH (aq)  Fe(OH)3 (s) + 3 NaCl (aq)
If L of the iron (III) chloride solution reacts with an unlimited NaOH, how many grams of precipitate will be made?
0.025 L FeCl3 x 0.2 mol FeCl3 x 1 mol Fe(OH)3 x g Fe(OH)3 = g Fe(OH)3
1.0 L mol FeCl mol Fe(OH)3