1. Solution STOICHIOMETRY!!! (Hooray!)

2. Joke… (so you’d better laugh!) Everyone has problems… But Chemistry students have SOLUTIONS!!!

3. Stoich. again?!? Lots of important reactions happen in solution! – SO… we need to know how to do stoich calculations ! – Good News: Most of the steps are the same!

4. Steps for Solution Stoich 1.Write the balanced equation (Hint: if ions are involved, writing the net ionic equation will be very helpful!) 2.Start with the given volume 3.Use molarity (mol solute/L solution) to convert volume  moles 4.MOLE RATIO!!! 5.Convert from moles to the desired units using molarity or molar mass

5. Example: When Ba(NO 3 ) 2 and K 2 CrO 4 react in aqueous solution, the yellow solid BaCrO 4 is formed. – Calculate the mass of BaCrO 4 that forms when excess solid Ba(NO 3 ) 2 is dissolved in 265 mL of 0.0100 M K 2 CrO 4 solution. Net Ionic Equation: Ba 2+ (aq) + CrO 4 2- (aq)  BaCrO 4 (s)

6. Convert given volume to moles: 265 mL K 2 CrO 4 x 1L x 0.0100mol = 1000mL 1 L 2.65x10 -3 mol K 2 CrO 4

7. Mole Ratio(s): 2.65x10 -3 mol K 2 CrO 4 x 1 mol CrO 4 2- x 1 mol BaCrO 4 1 mol K 2 CrO 4 1 mol CrO 4 2- 2.65x10 -3 mol BaCrO 4

8. Convert to desired units (grams for this problem!) 2.65x10 -3 mol BaCrO 4 x 253.3 g BaCrO 4 = 1 mol BaCrO 4 0.671 g BaCrO 4