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Unit 12: Stoichiometry Chapter 12 Chemistry 1K Cypress Creek High School Reminders… • The mole is a counting unit – 1 mole = 6.02 x 1023 particles (atoms, molecules, formula units) • Molar mass is the mass of 1 mole – Measured in grams – Represented by the atomic mass shown on the periodic table • Example: 1 mole of water (H2O) is about 18 grams. It contains 2 moles of hydrogen atoms and 1 mole of oxygen atoms. Mole Map What is Stoichiometry? • Using the methods of stoichiometry, we can measure the amounts of substances involved in chemical reactions and relate them to one another. – Pronounced stoy-kee-AW-muh-tree – From the Greek words stoikheion meaning element and metria meaning measure Stoichiometric Calculations • There are several basic stoichiometric calculations: – mole-to-energy conversions – mole-to-mole conversions – mole-to-mass / mass-to-mole conversions – mass-to-mass conversions – mole-to-volume conversions • All stoichiometric calculations are based on a balanced equation and mole ratios. Mole Ratios • Mole ratios indicate the molar relationship between two chemicals in an equation – Use the coefficients (not subscripts) to make ratios – Reduce to lowest whole-number ratio • Example: Water Formation – 2H2 + O2 2H2O • What is the ratio of hydrogen to oxygen? 2:1 • What is the ratio of oxygen to water? 1:2 • What is the ratio of water to hydrogen? 1:1 Mole-to-Mole Conversions • The following reaction shows table salt production. How many moles of sodium chloride are produced from 0.02 moles of chlorine? Solve through ratios: X mol NaCl = 2 mol NaCl 0.02 mol Cl2 1 mol Cl2 (cross multiply and solve) = 0.04 mol NaCl Mole-to-Energy Conversions • In the decomposition of baking soda, NaHCO3(s) + 1,800 kJCO2(g) + NaOH(s), how much energy is required if there is 3 moles of NaOH produced? Solve through ratios: X kJ = 3 mol NaOH 1800kJ 1 mol NaOH (cross multiply and solve) = 5400kJ Mole-to-Volume Conversions • Given 1 mol of O2, what volume of NO is needed in the reaction: NO + O2 NO3? Solve through ratios: X L NO = 1mol O2 1(22.4)L H2 1 mol O2 (cross multiply and solve) = 22.4L NO Volume-to-Volume Conversions • N2 + 3H2 2NH3, what volume of hydrogen is necessary to react with 5L of nitrogen to produce ammonia? Solve through ratios: X L H2 = 5L N2 3(22.4)L H2 1(22.4)L N2 (cross multiply and solve) = 15L Mole-to-Mass Conversions • The following reaction shows photosynthesis. How many grams of glucose are produced when 24 moles of carbon dioxide reacts with water? Solve through ratios: X g C6H12O6 = 24 mol CO2 180 g C6H12O6 (cross multiply and solve) = 720 g C6H12O6 6 mol CO2 Mass-to-Mole Conversions • The following reaction shows photosynthesis. How many moles of glucose are produced when 132 grams of carbon dioxide reacts with water? Solve through ratios: X mol C6H12O6 = 1 mol C6H12O6 132 g CO2 6(44 g CO2) (cross multiply and solve) = 0.5 mol C6H12O6 Mass-to-Mass Conversions • The following reaction shows the production of ammonia. How many grams of nitrogen are required to produce 85 grams of ammonia? Solve through ratios: X g N2 = 1(28 g N2) 85 g NH3 2(17 g NH3) (cross multiply and solve) = 70 g N2 Limiting Reactant • In the real world, reactants are not present in the exact mole ratio described by the balanced equation. • This means that one of the reactants will be used up before the other one. – The limiting reactant is used up first and restricts (stops) the reaction – The excess reactant(s) remain after the reaction stops • Example: If you placed a single drop of water on an Alka-Seltzer tablet… – What would be the limiting reactant? – Excess reactant? – How could you switch the roles of these two reactants? Limiting Reactant • Experiment: Equal amounts of copper (II) chloride are combined with varying amounts of aluminum to form copper and aluminum chloride. Which pairing will react completely? • 2 Al(s) + 3 CuCl2(aq) 2 AlCl3(aq) + 3 Cu(s) A B C 1:1 mass 1:1 mole 2:3 mass D Al:CuCl2 ratio 2:3 mole Limiting Reactant • Results: In reactions A, B, & C, aluminum is still visible the excess reactant. Copper (II) chloride is the limiting reactant. Reaction D, which follows the correct ratio in the balanced equation, reacted completely. • 2 Al(s) + 3 CuCl2(aq) 2 AlCl3(aq) + 3 Cu(s) A B C D Al:CuCl2 ratio 1:1 mass 1:1 mole 2:3 mass 2:3 mole Limiting Reactant Practice • Diagram A represents reactant particles, and Diagram B represents product particles in a fictitious chemical reaction. Write the balanced Diagram A Diagram B equation for this reaction. A Z C A Z C Z A A A A A A A Z C Z A Z A A Z C Z A A What is the limiting reactant? Excess? Limiting Reactant Practice • The three particle diagrams to the right represent the burning of methane. In which diagram is the limiting reactant oxygen gas? CH4 + 2O2 CO2 + 2 H2O A B C Limiting Reactant Calculations Method 1 • In a reaction, 80 grams of sodium hydroxide is combined with 60 grams of sulfuric acid. What is the limiting reactant? • Convert one of the masses of a given reactants into the mass of the other: Xg NaOH = 60g H2SO4 = 48.98g NaOH 2(40g NaOH) 1(98g H2SO4) • Compare the mass calculated to the mass given in the problem – 48.98g of NaOH is needed to react with 60 g of H2SO4 – You are given 80g of NaOH, therefore there is more NaOH than required to react with 60g of H2SO4 – Sulfuric acid is the limiting reactant Limiting Reactant Calculations Method 2 • In a reaction, 80 grams of sodium hydroxide is combined with 60 grams of sulfuric acid. What is the limiting reactant? • Calculate the moles of one of the product from each reactant (use H2O if present): X mol H2O = 80 g NaOH 2 mol H2O 2(40g NaOH) X = 2 mol H2O x mol H2O = 60 g H2SO4 2 mol H2O 1(98g H2SO4) x = 1.22 mol H2O • Compare the 2 molar amounts – Sulfuric acid produces less water – Sulfuric acid is the limiting reactant Percent Yield • Percent yield compares the amount of product collected in an experiment (actual) to the amount anticipated according to calculations (theoretical). • Actual mass will always be less than theoretical mass due to human error actual mass of product x 100 % yield = theoretical mass of product Percent Yield Calculations • Methanol, (CH3OH), was used as one of several embalming fluids in ancient Egypt. In an experiment, you used 25 grams of CO2 with excess hydrogen gas to produce 15 grams of methanol. What is the percent yield? • Calculate theoretical yield X g CH3OH = 25 g CO2 = 18.18 g 1( 32g CH3OH) 1(44 g CO2) • Calculate percent yield 15 g CH3OH x 100 = 82.5% yield 18.18 g CH3OH