WE BE COOKING…. Stoichiometry. Deals with the chemistry of quantifying chemical reactions through the use of:  Balancing equations  Moles (mol)  Molarity.

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Presentation transcript:

WE BE COOKING…. Stoichiometry

Deals with the chemistry of quantifying chemical reactions through the use of:  Balancing equations  Moles (mol)  Molarity (mol/L)  Mass (grams)

Cooking… A recipe is used to make a dish…let’s say brownies ½ cup butter 2 oz of chocolate 1 cp sugar 2 eggs 1 tsp vanilla 2/3 cp flour ½ tsp baking powder ¼ tsp salt 24 Brownies

What if I want 36 Brownies The ingredients don’t change but the amount does proportionally… ½ cup butter 2 oz of chocolate 1 cp sugar 2 eggs 1 tsp vanilla 2/3 cp flour ½ tsp baking powder ¼ tsp salt ¾ cup butter 3 oz of chocolate 1.5 cp sugar 3 eggs 1.5 tsp vanilla 1 cp flour ¾ tsp baking powder 1/8 tsp salt x by 1.5

Recipes are just like chemical equations 2Mg(s) + O 2 (g)  2MgO(s) 2 mol of Mg + 1 mol of O 2 yields 2 moles of MgO If I want 6 moles of MgO…I need 6 mol of Mg and 3 moles of O 2. Proportions… Wait…wait…proportion? That means conversion!! What…DA?

Coefficients are Conversion Factors 2KClO 3 (s)  2KCl(s) + 3O 2 (g) Conversions: 2 mol KClO 3 = 2 mol KCl 2 mol KClO 3 = 3 mol O 2 2 mol KCl = 3 mol O 2 Question: How many moles of oxygen gas is produced if there is 4.56 moles of KClO 3 present? 4.56 mol KClO 3 mol KClO 3 mol O Coefficients from balanced equation = 6.84 mol of O 2

So what does this mean… From previous knowledge:  Liters to moles (Molarity)  Mass to moles (molar mass) Now…  Moles to moles (using coefficients) We can not only work with one particular compound/molecule/atom…now we can work with multiple compounds/molecules/atoms All aboard to moleland!!

Let’s put this knowledge into action.. 2KClO 3 (s)  2KCl(s) + 3O 2 (g) How many grams of KCl is produced if 6.89 grams of KClO 3 are decomposed? 6.89 g KClO 3 g of KClO 3 mol of KClO Molar mass of KClO 3 1 mol KClO 3 mol KCl mol of KCl g of KCl Molar mass of KCl = 4.19 g of KCl Coefficients

Review Steps for Stoiching… Write and balancing chemical reaction Start with given (mass, moles, or volume) Convert to moles (if mass and volume is given)  Moleland!! Convert to moles of another compound/atom Convert to final unit (moles/grams/volume)

0.100 mol AgNO 3 Working with molarity Cu(s) + 2AgNO 3 (aq)  2Ag(s) + Cu(NO 3 ) 2 (aq) How many grams of copper will be required to completely replace silver from 208 mL of M solution of AgNO 3 ? 208 mL 1 mol Cu 1000 mL 1 L 2 mol AgNO 3 1 mol Cu g Cu Molar mass of Cu = g of Cu Coefficients 1 L

Reasons for Stoichiometry Quantify how much you need to start with  Without throwing in any amount willy nilly Quantify how much you make  Determine how baller you are Quantify how much you have left over

Lead(II) nitrate solution reacts with 27.5 mL of 3.00 M carbonic acid, H 2 CO 3 (aq), how many grams of lead(II) carbonate is produced? 3.00 mol H 2 CO 3 More Practice 27.5 mL 1 mol PbCO mL 1 L 1 mol H 2 CO 3 1 mol PbCO g PbCO 3 Molar mass of PbCO 3 = 22.0 g of PbCO 3 Coefficients 1 L Pb(NO 3 ) 2 (aq) + H 2 CO 3 (aq)  PbCO 3 (s) + 2HNO 3 (aq)

Solid limestone, calcium carbonate, is heated to produced solid CaO and carbon dioxide gas. How much limestone is required to produce 10.0 grams of calcium oxide? 1 mol CaO More Practice 10.0 g CaO 1 mol CaCO 3 1 mol CaO 1 mol CaCO g CaCO 3 Molar mass of CaCO 3 = 17.8 g of CaCO 3 Coefficients g CaO CaCO 3  CaO(s) + CO 2 (g) Molar mass of CaO