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Starter S-125. 1.2 moles are NaC 2 H 3 O 2 are used in a reaction. How many grams is that?. Stoichiometry. Chapter 12. 12.1 The Arithmetic of Equations. Chapter 12. 12.1 The Arithmetic of Equations. The basis for solving stoichiometry problems is a balanced chemical reaction
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Starter S-125 1.2 moles are NaC2H3O2 are used in a reaction. How many grams is that?
Stoichiometry Chapter 12
12.1 The Arithmetic of Equations Chapter 12
12.1 The Arithmetic of Equations The basis for solving stoichiometry problems is a balanced chemical reaction A balanced reaction is used to calculate How much reactant is needed How much product is produced
12.1 The Arithmetic of Equations Stoichiometry – the calculation of quantities This reaction produces Ammonia which is used in fertilizers Balanced reactions are usually used to calculate grams of product or reactant
12.1 The Arithmetic of Equations Analysis of the reaction Atoms – 2 atoms of nitrogen combine with 6 atoms of hydrogen – product is 2 nitrogen and 6 hydrogen Molecules – 1 molecule of nitrogen gas combines with 3 molecules of hydrogen gas to produce 2 molecules of ammonia
12.1 The Arithmetic of Equations Analysis of the reaction Most useful Moles – 1 mole of nitrogen gas reacts with 3 moles of hydrogen gas to produce 2 moles of ammonia
12.1 The Arithmetic of Equations Analysis of the reaction Mass is always conserved in a chemical reaction
12.1 The Arithmetic of Equations Analysis of the reaction The key moles can be converted to grams grams can be converted to moles Volume – remember that one mole of gas at STP is 22.4L
12.1 The Arithmetic of Equations Analysis of the reaction The key moles can be converted to grams grams can be converted to moles Volume – remember that one mole of gas at STP is 22.4L
12.1 The Arithmetic of Equations Analysis of the reaction So if we start with 50g of N2, how many moles do we have How many moles of NH3 would be produced? For every 1 mole of N2, 2 moles of NH3
12.1 The Arithmetic of Equations Analysis of the reaction How many grams of ammonia are produced
Starter S-126 A. Balance the following reactions H2SO4 + KOH K2SO4 + H2O • What is the molar mass of the sulfuric acid. • If 20.0g of sulfuric acid is used, how many moles is that? • How many moles of water would be produced?
12.2 Chemical Calculations Chapter 12
12.2 Chemical Calculations 12.2 Chemical Calculations Mole Ratios Come from balanced chemical reactions Conversion factors derived from the coefficients in the balanced reactions
12.2 Chemical Calculations Mole-Mole Calculations It is possible to convert from one quantity in a balanced reaction to another using mole ratios For example – if 3.7 moles of sulfur dioxide is produced, how many moles of oxygen were used?
12.2 Chemical Calculations Mass-Mass Calculations Three steps • Convert given mass values into mole values. • Use a mole ratio to convert to the moles that the question requests • Convert this mole quantity to a mass value
12.2 Chemical Calculations Mass-Mass Calculations How many grams of Oxygen are needed to produce 30.0g of Sulfur Dioxide?
12.2 Chemical Calculations Volume-Mass Calculations Same steps, but volume is converted to moles, or moles to volume Example: If 4.0L of nitrogen monoxide reacts, how many grams of oxygen are used?
Starter S-127 A. Balance the following reactions Al(NO3)3 + Na2SO4 Al2(SO4)3+NaNO3 B. If 50.0g of Aluminum Nitrate reacts, how many grams of Aluminum Sulfate are produced?
12.3 Limiting Reagent and Percent Yield Chapter 12
12.3 Limiting Reagent and Percent Yield Limiting Reagent – the reactant that runs out first Amounts of both reactants are given Example: 80.0g Copper, 25.0g Sulfur • Calculate how many moles would each reactant produce Copper
12.3 Limiting Reagent and Percent Yield Limiting Reagent – the reactant that runs out first Amounts of both reactants are given Example: 80.0g Copper, 25.0g Sulfur • Calculate how many moles would each reactant produce Sulfur
12.3 Limiting Reagent and Percent Yield Limiting Reagent – the reactant that runs out first From the reactions Copper would produce 0.630 mol Cu2S Sulfur would produce 0.780 mol Cu2S That means copper will run out first – it is the limiting reagent Sulfur would be the excess reagent
12.3 Limiting Reagent and Percent Yield One for you now • Balance the reaction • If 75.6g C2H4 reacts with 100.8g O2 – what is the limiting reagent? Oxygen • How many moles of water are produced? 2.10 mol H2O 4. How many grams of water are produced?
Starter S-132 A. Balance the following reactions SiO2 + C SiC + CO B. 35 grams of silicon dioxide reacts with 10.0g of Carbon, how much carbon monoxide is formed?
12.3 Limiting Reagent and Percent Yield Reactions rarely produce as much product as is predicted -reactants can be impure -reactions may not go to completion -may compete with smaller “side” reactions In some reactions as little as 60% yield is considered a good result
12.3 Limiting Reagent and Percent Yield Yield – how much product is produced Theoretical Yield – the value calculated using stoichiometry Actual Yield – the amount of product that actually forms
12.3 Limiting Reagent and Percent Yield Percent Yield – a ratio of actual to theoretical yield This number must be 100% or less In lab the actual yield is the result you get On a test, it will be a number that is given to you The theoretical yield is calculated using limiting reactants
12.3 Limiting Reagent and Percent Yield Example: What is the theoretical yield of Calcium Oxide if 24.8g of Calcium Carbonate decomposes in the following reaction. Balanced
12.3 Limiting Reagent and Percent Yield Example: What is the percent yield if actual yield is 9.6g?
12.3 Limiting Reagent and Percent Yield Example: What is the theoretical yield if 15.0g of nitrogen reacts with 15.0g of hydrogen in the following reaction? Balance
12.3 Limiting Reagent and Percent Yield Example: If the actual yield is 10.5g of NH3 what is the percent yield?