1. Solutions Review
HW study for test.

2. Solubility Curves How are solubility and temperature related?
What is the molarity of 250. mL potassium nitrate at 70°C and how many moles of nitrate are present?
Molarity = 12.9 M KNO3
3.21 moles NO31-

3. If you want to make mL M calcium chloride solution, how many moles of CaCl2 will you need? If you dilute this solution to 750. mL what is the new concentration?
mol = M∙L = (0.100 M)(0.250 L) = mol CaCl2
M1V1 = M2V2
M2 = M1V1 = (0.1M) (250.0 mL) = M CaCl2
V mL

4. Stoichiometry with Limiting Reactants
Calculate the mass of Ag2CO3(s) produced by mixing 125 mL of M Na2CO3(aq) and 75.0 mL of M AgNO3(aq).
What we know:
Na2CO3(aq) + 2 AgNO3(s)  Ag2CO3(s) NaNO3(aq)
0.315 M M ? mass
125 mL mL

5. Na2CO3(aq) + 2 AgNO3(s)  Ag2CO3(s) + 2 NaNO3(aq)
0.315 M M ? mass
125 mL mL
OPTIONAL -You can also think in terms of a Net Ionic Equation:
CO32− (aq) Ag+ (aq)  Ag2CO3(s)
0.315 M M ? Mass
125 mL mL
Now convert both reactants to moles: (mol = M ∙ L)
mol CO32- = (0.315 M AgNO3 ) (0.125 L) = mol Na2CO3
mol Ag + = (0.155 M Na2CO3) (0.075L) = mol AgNO3

6. Change moles of AgNO3 and Na2CO3 to moles of product to determine the limiting reactant
Na2CO3(aq) + 2 AgNO3(s)  Ag2CO3(s) NaNO3(aq)
mol mol
Calculations:
mol AgNO3 1mol Ag2CO g Ag2CO3
2mol AgNO mol Ag2CO3
= 1.60 g Ag2CO3
mol Na2CO3 1mol Ag2CO g Ag2CO3
mol Na2CO mol Ag2CO3
= 10.9 g Ag2CO3
AgNO3 is Limiting Reactant
( )
( )
( )
( )