Solutions Review HW study for test..

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Presentation transcript:

Solutions Review HW study for test.

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-

If you want to make 250.0 mL 0.100 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) = 0.0250 mol CaCl2 M1V1 = M2V2 M2 = M1V1 = (0.1M) (250.0 mL) = 0.0333 M CaCl2 V2 750 mL

Stoichiometry with Limiting Reactants Calculate the mass of Ag2CO3(s) produced by mixing 125 mL of 0.315 M Na2CO3(aq) and 75.0 mL of 0.155 M AgNO3(aq). What we know: Na2CO3(aq) + 2 AgNO3(s)  Ag2CO3(s) + 2 NaNO3(aq) 0.315 M 0.155 M ? mass 125 mL 75.0 mL

Na2CO3(aq) + 2 AgNO3(s)  Ag2CO3(s) + 2 NaNO3(aq) 0.315 M 0.155 M ? mass 125 mL 75.0 mL OPTIONAL -You can also think in terms of a Net Ionic Equation: CO32− (aq) + 2 Ag+ (aq)  Ag2CO3(s) 0.315 M 0.155 M ? Mass 125 mL 75.0 mL Now convert both reactants to moles: (mol = M ∙ L) mol CO32- = (0.315 M AgNO3 ) (0.125 L) = 0.0394 mol Na2CO3 mol Ag + = (0.155 M Na2CO3) (0.075L) = 0.0116 mol AgNO3

Change moles of AgNO3 and Na2CO3 to moles of product to determine the limiting reactant Na2CO3(aq) + 2 AgNO3(s)  Ag2CO3(s) + 2 NaNO3(aq) 0.0394 mol 0.0116 mol Calculations: 0.0116 mol AgNO3 1mol Ag2CO3 275.75g Ag2CO3 2mol AgNO3 1mol Ag2CO3 = 1.60 g Ag2CO3 0.0394 mol Na2CO3 1mol Ag2CO3 275.75g Ag2CO3 mol Na2CO3 1mol Ag2CO3 = 10.9 g Ag2CO3 AgNO3 is Limiting Reactant ( ) ( ) ( ) ( )