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Transcript
Chapter 3: Stoichiometry ● “measuring elements” ● Must account for ALL atoms in a chemical reaction → + 2 H2 + O2 → 2 H 2O Chapter 3: Stoichiometry → + 2 CO + + 2 CO2 → + CH4 → O2 4 Cl2 → + CCl4 + 4 HCl Chapter 3: Stoichiometry C 2H 4 3 O2 + 2 Al + 2 NH4NO3 H2O 6 HCl → 2 N2 → 2CO2 → 2 AlCl3 + O2 + + + 4 3 H2 2 H 2O Chapter 3: Stoichiometry Three basic reaction types: ● Combination Reactions ● Decomposition Reactions ● Combustions (in air) Chapter 3: Stoichiometry Combination Reactions Two or more reactants combine to form a single product C (s) + O2 (g) → CO2 (g) Decomposition Reactions A single reactant breaks into two or more products 2 KClO3 (s) → 2 KCl (s) + 3 O2 (g) Chapter 3: Stoichiometry Combustions in Air = reactions with oxygen Write the balanced reaction equation for the combustion of magnesium to magnesium oxide: Chapter 3: Stoichiometry Combustions of Hydrocarbons in Air = reactions with oxygen to form carbon dioxide and water (complete combustion) Write the balanced reaction equation for the combustion of C2H4 gas C2H4 (g) Chapter 3: Stoichiometry C2H4 (g) + 3 O2 (g) → → + 2 CO2 (g) + 2 H2O (g) + How many C2H4 molecules are in the flask? If you know the weight of one molecule of C2H4 and the total weight of gas in the flask, you can calculate the number of molecules in the flask ● Chapter 3: Stoichiometry Molecular weight / Formula weight: => sum of all atomic weights in molecular formula for molecular compounds MW of C2H4 FW of Mg(OH)2 for ionic compounds Chapter 3: Stoichiometry Ca(NO3)2 Type of compound: Ions: Total number of oxygen atoms: Name: Chapter 3: Stoichiometry Ca(NO3)2 Percentage of oxygen, by mass: (1) total mass of Ca(NO3 )2 in amu (2) mass of oxygen in compound, in amu (3) percentage of oxygen Chapter 3: Stoichiometry Number of individual molecules are difficult to deal with => definition of a “package” of molecules or particles . ... .. .. . . . . . . .. ... .... ..... .. . .. . . .. 1 dozen eggs = 12 individual eggs 1 soda sixpack = 6 individual bottles of soda 1 mole of molecules = 6.02 x 1023 individual molecules Avogadro's Number Chapter 3: Stoichiometry 1 dozen eggs = 12 individual eggs How many moles of eggs are in an egg carton that holds a dozen eggs? 12individual eggs × Chapter 3: Stoichiometry 1 dozen eggs = 12 individual eggs How many moles of eggs are in an egg carton that holds a dozen eggs? Chapter 3: Stoichiometry Ca(NO3)2 How many moles of oxygen are in 2.4 moles of Ca(NO3)2 ? Chapter 3: Stoichiometry Molar Mass = mass of one mole of a substance in grams FW or MW of substance in amu's = mass of 1mole of substance in grams FW of Ca(NO3)2 = 164.1 amu Molar Mass of Ca(NO3)2 = 164.1 g/mol MW of O2 = 2 x 16.0 amu = 32 amu Molar Mass of O2 = 32 g/mol Chapter 3: Stoichiometry What is the mass in grams of 0.527 moles of CH3OH? (1) determine molar mass of CH3OH 32 g/mol (2) use MM to convert moles into grams 16.9 g CH3OH Chapter 3: Stoichiometry mol → ← Molar Mass gram How many hydrogen atoms are in 4.5 g of CH3OH? = 0.14 mol CH3OH = 0.56 mol H atoms = 3.4 x 1023 H atoms Chapter 3: Stoichiometry Quantitative Information from Balanced Equations How many grams of CO2 would be produced by the combustion of 2 moles of CO? 2 CO + → O2 2 CO2 → + 2 moles CO + 1 mole O2 2 x 28 g + 32 g → → 2 moles CO2 2 x 44 g = 88 g Chapter 3: Stoichiometry Quantitative Information from Balanced Equations 2 CO + CO2 O2 → 2 You can write a series of stoichiometric factors for this reaction: 2mol CO 1 mol O2 1 mol O2 2mol CO 1 mol O2 2 mol CO2 Chapter 3: Stoichiometry How many grams of H2O are formed from the complete combustion of 2 g of C2H4? 18 g/mol 28 g/mol C2H4 (g) 1 mol + 3 O2 (g) → 2 CO2 (g) + 2 H2O (g) 2 mol = 2.6 g H2O Chapter 3: Stoichiometry Summary 1) determine equation for the reaction 2) balance equation 3) formulate problem: how much of A => gets converted into how much of B 4) determine MW/FW of substances involved 5) determine stoichiometric factors from balanced equation Chapter 3: Stoichiometry Limiting Reactants + + 2 of these will be left over “in excess” Limiting Reactant Chapter 3: Stoichiometry Limiting Reactants Limiting Reactant - limits the amount of product that can be formed - reacts completely (disappears during the reaction) - other reactants will be left over, i.e. in excess Chapter 3: Stoichiometry Limiting Reactants N2 Available: 3 mol N2 + , 3 H2 → 2 NH3 6 mol H2 How much H2 would we need to completely react 3 mol N2: = 9 mol H2 How much NH3 can we form with the available reagents? = 4 mol NH3 Chapter 3: Stoichiometry Limiting Reactants N2 Available: 3 mol N2 + , 3 H2 → 2 NH3 6 mol H2 How much N2 is left over (in excess)? = 1 mol N2 Chapter 3: Stoichiometry Limiting Reactants 2 Al + 3 Cl2 → 2 AlCl3 Available: 0.5 mol Al , 2.5 mol Cl2 How much Cl2 would we need to completely react 0.5 mol Al: = 0.75 mol Cl2 How much AlCl3 can we form with the available reagents? = 0.5 mol AlCl3 Chapter 3: Stoichiometry Theoretical Yield 2 Al + 3 Cl2 → 2 AlCl3 Available: 0.5 mol Al , 2.5 mol Cl2 What mass do 0.5 mol AlCl3 correspond to? = 67 g AlCl3 The maximum mass of product that can be formed is the theoretical yield Chapter 3: Stoichiometry Theoretical Yield 2 Al + 3 Cl2 → 2 AlCl3 Available: 0.5 mol Al , 2.5 mol Cl2 Fritz does the reaction with the available reagents he only ends up with 34g. What is the % yield of the reaction? % yield = actual yield × 100 % theoretical yield % yield 34 g 100 % = 51 % 67 g Chapter 3: Stoichiometry Summary Determine availabe quantity of reactants in moles Determine % yield of the reaction Determine if one of the reactants is a limiting reactant Compare with actual amount of product recovered (actual yield) Determine the maximum # of moles of product that can be formed Convert into grams of product (theoretical yield) Chapter 3: Stoichiometry Consider the combustion of methanol: 2 CH3OH + 3 O2 → 2 CO2 + 4 H2O What is the theoretical yield of water if 44 g of methanol are reacted with 128 g of oxygen? 1.4 mol CH 3OH 4.0 mol O2 Chapter 3: Stoichiometry 2 CH3OH + 3 O2 available (initial) moles 1.4 mol → 2 CO2 + 4 H2O 4.0 mol O2 = 2.1 mol O2 Chapter 3: Stoichiometry 2 CH3OH + 3 O2 → 2 CO2 + 4 H2O (3) how many product moles can be formed with limiting reactant ? = 1.4 mol CO2 = 2.8 mol H2O Chapter 3: Stoichiometry 2 CH3OH + 3 O2 → 2 CO2 + 4 H2O (4) What is the mass of H2O formed (theoretical yield )? 50g H2O