Stoichiometry Section 12.1

What is Stoichiometry? Study of quantitative relationships between amounts of reactants used and products formed. Based on the Law of Conservation of Mass (Mass of the Reactants = Mass of the Products) Ex. 4 Fe (s) + 3 O 2 (g)  2 Fe2O3 (s)

Mole – Mass Relationships Coefficients in a balanced equation can be interpreted in terms of representative particles and also by numbers of moles of particles. We can also use what we know about the conversion factor that relates mass and number of moles to find the mass of either the reactants or products.

Mole – Mass Relationships Ex. 4 Fe (s) + 3 O2 (g)  2 Fe2O3 (s) 4 mol Fe x 55.85 g Fe = 223.4 g Fe 1 mol Fe 3 mol O2 x 32.00 g O2 = 96.0 g O2 1 mol O2 Mass of Reactants = 319.4 g 2 mol Fe2O3 x 158.7 g Fe203 = 319.4 g 1 mol Fe203 Mass of Products = 319.4 g

Ex. 2 Al (s) + 3 Br2 (l)  2 AlBr3 (s) Mole Ratios Ratio between the numbers of moles of any two substances in a balanced chemical equation. Ex. 2 Al (s) + 3 Br2 (l)  2 AlBr3 (s) 2 mol Al and _2 mol Al_ 3 mol Br2 and _3 mol Br2_ 3 mol Br2 2 mol AlBr3 2 mol Al 2 mol AlBr3 2 mol AlBr3 and 2 mol AlBr3 2 mol Al 3 mol Br2

Mole Ratio Practice What mole ratios can be written for the decomposition of potassium chlorate? 2 KClO3 (s)  2 KCl (s) + 3 O2 (g) Answer: 2 mol KClO3 and 2 mol KClO3 2 mol KCl 3 mol O2 _2 mol KCl_ and _2 mol KCl_ 2 mol KClO3 3 mol O2 __3 mol O2_ and _3 mol O2 2 mol KClO3 2 mol KCl

Mole – Mole Conversions All stoichiometric calculations begin with a balanced equation. Mole ratios based on the balanced chemical equation are also needed. A limiting reactant is a reactant that limits the extent of the reaction and determines the amount of product. This is the reactant you have the least amount of. Calculation: Moles of known x moles of unknown = moles of unknown moles of known

Mole – Mole Conversion Practice 2 K (s) + 2 H2O (l)  2 KOH (aq) + H2 (g) How can you determine the number of moles of hydrogen produced when .0400 mole of potassium is used? Write the balanced equation Given = .0400 mole K Unknown = moles of H2 3. Write the mole ratio: 1 mole H2 2 mole K Convert using: moles of known x moles of unknown moles of known Answer: .0400 x 1 mole H2 = .0200 mol H2 2 moles K

Mole – Mole Relationships in Equations Balance the equation. Coefficients = number of moles of each substance. Use “x” for the “how many” compound. Use ratio of moles given in problem to actual moles in equation. Set up ratio. Solve.

Mole – Mole Relationships in Equations Example 2 H2 + O2  H2O How many moles of water can be produced with 6 moles of hydrogen? Equation is balanced. There are 2 moles of hydrogen, 1 mole oxygen, and 2 moles water. 6 H2 = x H2O 6 = x and 2x = 12 2 H2 2 H2O 2 2 x = 6 4. X = 6 moles water

Stoichiometry Practice I (Mole-Mole Conversions) 2 H2 + O2  2 H2O 1. How many moles of water can be produced with 6 moles of hydrogen? 2. How many moles of oxygen would be required to fully react with 8 moles of hydrogen? 3. How many moles of water can be produced with 4 moles of oxygen? 4. How many moles of hydrogen would be required to produced 10 moles of water? 5. How many moles of oxygen would be needed to produce 20 moles of water?

Mole to Mass Relationships in Equations Balance the equation. Coefficients = # of moles of each substance. Use “x” for the “how many” compound. Use ratio of moles given in problem to actual moles in equation. Set up ratio and solve for “x”. Multiply your answer by the molar mass of the element or compound you are trying to find. (Example: Stoichiometry Practice II, # 4)

Mass to Mass Relationships in Equations Balance the equation. Find the molar mass of each of the reactants and products in the equation. Use “x” for the “how many” compound. Use ratio of mass given in problem to actual mass in equation. Set up the ratio and solve for “x”. (Example: Stoichiometry Practice II, # 6)