Chem 9 15 17 - Lecture notes 11 PDF

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Calculating Amounts of Reactions and Products
Dr. Gounev...

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Calculating Amounts of REactants and Products Stoichiometric Equivalents ● Balanced chemical equations contain definite stoichiometric relations between reactants and products → stoichiometric mole ratios ● Example: ○ 2H2 + O2 → 2H2O ○ 2 mol H2 → 1 mol O2 ○ 2 mol H2 → 2 mol H2O ○ 1 mol O2 → 2 mol H2O ○ 1 mol O2 / 2 mol H2 ○ 2 mol H2O / 2 mol H2 ● Mole to mole conversions ○ Conversion method ○ The mole ratios are used as conversion factors ○ (mol given)X(mole ratio)=(mol required) ○ Example: ■ Determine the number of moles of water produced from 3.4 mol O2 ■ Balanced equation: 2H2 +O2 → 2H2O ■ Mole ratio (conversion factor) : [2 mol H2O/1mol O2] ■ 3.4 mol O2 X ( 2mol H2O / 1 mol O2) = 6.8 mol ○ Stoichiometric conversion factors are reaction specific ○ Example: ■ Calculate the amount of O2 needed to produce 3.5 mol H2O by combustion of methane (CH4) ● Balanced equation= CH4+2O2 → CO2 +2H2O ● Mole ratio ○ 2 mol O2 → 2 mol H2O ● 3.5 mol H2O X ( 2 mol O2 / 2 mol H2O) = 3.5 mol O2 Mass to mass calculations: ● Example: ○ Calculate the mass of oxygen needed to completely burn 5.4 kg of Butane (C4H10) ■ 2C4H10 +13O2 → 8CO2 +10H2O ■ Mole ratio ● 13 mol O2 per 2 mol C4H10 ■ Molar masses ● C4H10 → 58.1 g/mol ● O2 → 32.0 g/mol ■ 5.4kcC4H10 X (10^3 g C4H10 / 1 kg C4H10) X ( 1 mol C4H10 / 58.1 g C4H10) X (13 mol O2 / 2 mol C4H10) X (32.0 g O2 / 1 mol O2) = 1.9x10^4 g O2= 19 kg O2 Reaction Yield ● Theoretical yield- the maximum amount of product that can be expected to form a

given amount of reactant Actual yield- the actual amount of product isolated in a reaction ○ Actual yield < theoretical yield ○ Reasons for the difference between actual and theoretical yield ■ Incomplete reaction ■ Loss of products ● Some product can dissolve in solvent ● Some product stuck on walls ■ Side reactions ● Percentage yield- actual yield divided by theoretical yield times 100% ● Example: ○ Calculate the theoretical yield of carbon dioxide produced by the combustion of 25.og of propane (C2H8) in excess O ■ C3H8 +5O2 → 3CO2 + 4H2O ■ Mass to mass conversion ● 25.0 g C3H8 X ( 1 mol C3H8 / 44.09 g C3H8) X ( 3 mol CO2 / 1 mol C3H8) X (44.01 g CO2 / 1 mol CO2) = 74.9 g CO2 → theoretical yield ● Example (Percent Yield): ○ Calculate the percentage yield of carbon dioxide if the combustion of 25.0 g propane in excess oxygen yields 48.5 g carbon dioxide ■ 48.5 g CO2 actually yields this much ■ Theoretical yield (from previous example) 74.9 g CO2 ■ Side reaction ( consumes some of the propane) ● 2C3H8 + 7O2 → 6CO + 8H2O ■ 48.5g CO2 / 74.9 g CO2 X 100% = 63.8% Limiting Reactants: ● Reactants present in equivalent amounts ○ All reactants are consumed at the same time ● Non Equivalent amounts of reactants ○ One reactant, called limiting reactant, is consumed before the others ○ The other reactants are in excess ● Limiting reactant ○ The reaction stops when the limiting reactant is consumed ○ Limits the maximum amount of product achievable (limits theoretical yield) ○ Stoichiometric calculations based on the limiting reactant give the lowest amount of product compared to calculations based on the other reactants ● Example: ○ Identify the limiting reactant in the reaction of 5.0 mol H2 with 3.0 mol N2 and determine the theoretical yield of NH3 in this reaction ■ 3H2 + N2 → 2NH3 ■ Take the theoretical yield of each reactant and take the smallest number as your limiting reactant ■ 3.0 mol N2 ( 2 mol NH3 / 1 mol N2) = 6.0 mol NH3 ●

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5.0 mol H2 X ( 2 mol NH3 / 3 mol H2) = 3.3 mol NH3 → limiting reactant and theoretical yield

Example: ○ Calculate the theoretical yield of HNO3 in the reaction of 28 g NO2 and 18 g H2O by the chemical equation: ■ 3NO2 + H2O → 2HNO3 + NO ■ Calculate the theoretical yield based on each of the reactants and chose the smaller result ● 18 g H2O X ( 1 mol H2O / 18.0 g H2O) X ( 2 mol HNO3 / 1 mol H2O) X ( 63.0 g HNO3 / 1 mol HNO3) = 130 g HNO3 ● 28 g NO2 X ( 1 mol NO2 / 46.0 g NO2) X ( 2 mol HNO3 / 3 mol NO2) X ( 63.0 g HNO3 / 1 mol HNO3 ) = 26 g HNO3 → theoretical yield...