2. Proportional Relationships StoichiometryStoichiometry –mass relationships between substances in a chemical reaction –based on the mole ratio Mole RatioMole Ratio –indicated by coefficients in a balanced equation 2 Mg + O 2  2 MgO Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

3. Stoichiometry Island Diagram Mass Particles Mole Mass Particles Known Unknown Substance A Substance B Stoichiometry Island Diagram 1 mole = molar mass (g) Use coefficients from balanced chemical equation 1 mole = 6.022 x 10 23 particles (atoms or molecules) 1 mole = 6.022 x 10 23 particles (atoms or molecules) 1 mole = molar mass (g)

4. ? Visualizing a Chemical Reaction Na + Cl 2 NaCl ___ mole Cl 2 ___ mole NaCl___ mole Na 2 105 2 5

5. Formation of Ammonia

6. Stoichiometry Steps 1. Write a balanced equation. 2. Identify known & unknown. 3. Line up conversion factors. –Mole ratio - moles  moles –Molar mass -moles  grams –Avogadro’s number - particles  moles Core step in all stoichiometry problems!! –Mole ratio - moles  moles 4. Check answer. Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

7. Stoichiometry Problems How many moles of KClO 3 must decompose in order to produce 9 moles of oxygen gas? 9 mol O 2 2 mol KClO 3 3 mol O 2 = 6 mol KClO 3 2KClO 3  2KCl + 3O 2 ? mol9 mol Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

8. Stoichiometry Problems How many grams of silver will be formed from 12.0 g copper? 12.0 g Cu 1 mol Cu 63.55 g Cu = 40.7 g Ag Cu + 2AgNO 3  2Ag + Cu(NO 3 ) 2 2 mol Ag 1 mol Cu 107.87 g Ag 1 mol Ag 12.0 g? g Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

9. Rocket Fuel The compound diborane (B 2 H 6 ) was at one time considered for use as a rocket fuel. How many grams of liquid oxygen would a rocket have to carry to burn 10 kg of diborane completely? (The products are B 2 O 3 and H 2 O). B 2 H 6 + O 2 Chemical equation Balanced chemical equation X = 34,286 g O2O2 10 kg x g x g O 2 = 10 kg B 2 H 6 1000 g B 2 H 6 1 kg B 2 H 6 28 g B 2 H 6 1 mol B 2 H 6 3 mol O 2 1 mol B 2 H 6 32 g O 2 1 mol O 2 B 2 O 3 + H 2 O 3 3 B 2 H 6 + O 2 B 2 O 3 + H 2 O

10. Water in Space In the space shuttle, the CO 2 that the crew exhales is removed from the air by a reaction within canisters of lithium hydroxide. On average, each astronaut exhales about 20.0 mol of CO 2 daily. What volume of water will be produced when this amount of CO 2 reacts with an excess of LiOH? (Hint: The density of water is about 1.00 g/mL.) CO 2 (g) + 2 LiOH(s)  Li 2 CO 3 (aq) + H 2 O(l) excess 20.0 mol x g X = 360 mL H2OH2O x mL H 2 O = 20.0 mol CO 2 1 mol H 2 O 1 mol CO 2 1 mol H 2 O 18 g H 2 O 1 mL H 2 O 1 g H 2 O 22.4 L H 2 O Water is NOT at STP!

11. Limiting Reactants Limiting ReactantLimiting Reactant –used up in a reaction –determines the amount of product Excess ReactantExcess Reactant –added to ensure that the other reactant is completely used up –cheaper & easier to recycle Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

12. Percent Yield calculated on paper measured in lab % yield = actual yield theoretical yield x 100

13. When 45.8 g of K 2 CO 3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K 2 CO 3 +  2KCl + H 2 O + CO 2 45.8 g 46.3 g actual yield excess 2HCl theoretical yield Theoretical yield x g KCl = 45.8 g K 2 CO 3 = 49.4 g KCl 1 mol K 2 CO 3 138 g K 2 CO 3 2 mol KCl 1 mol K 2 CO 3 46.3 g KCl % Yield = Actual Yield Theoretical Yield % Yield % Yield = 93.7% efficient 49.4 g KCl = x 100 74.5 g KCl 1 mol KCl 49.4 g ? g