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Stoichiometry Unit: Stoichiometry Topic: Intro to Stoichiometry Objectives: Day 1 of 4 • To learn how we go from moles of a reactant to moles of a product • To learn how to calculate between moles of reactants to moles of products Quickwrite Answer one of the questions below 1-2 sentences: • A sandwich recipe requires 2 pieces of bread, 2 slices of meat and 1 slice of cheese; you need to make 50 sandwiches, how many slices of bread, meat and cheese do you need???? Stoichiometry • Chemistry is really all about reactions • Reactions involve the rearrangement of atoms • The calculation of the quantities of chemical elements or compounds involved in chemical reactions is called Stoichiometry • It is the coefficients in the balanced chemical equation that enables us to determine just how much product forms • What we once called coefficients are now called moles!!!! What is Stoichiometry? quantities of • The calculation of ________ chemical elements or compounds involved in reactions chemical ________ • What we once called coefficients are now called _____ moles Answer Bank Reactants conversion moles Quantities Reactions products Stoichiometry • To explore this idea, consider a nonchemical analogy • A particular type of sandwich requires 2 pieces of bread, 3 slices of meat, and 1 slice of cheese • Or, you might represent this by: 2 pieces of bread + 3 slices of meat + 1 slice of cheese →1 sandwich Stoichiometry • Now, your boss sends you to the store to get enough ingredients to make 50 sandwiches • How do you figure out how much of each ingredient you need to buy? • You could multiply the previous equation by 50: 50(2 pieces of bread) + 50(3 slices of meat) + 50(1 slice of cheese) → 50(1 sandwich) 100 pieces of bread + 150 slices of meat + 50 slice of cheese → 50 sandwich Stoichiometry • Notice that the numbers 100:150:50 correspond to the ratio 2:3:1, which represents the original numbers of bread, meat and cheese • The equation for chemical reaction gives you the same type of information • It indicates the relative numbers of reactant and product molecules required for the reaction to take place Stoichiometry • To illustrate how this idea works, consider the reaction between gaseous carbon monoxide and hydrogen to produce liquid methanol • The reactants and products are: • Unbalanced: CO(g) + H2(g) → CH3OH(l) • Because atoms are just rearranged (not created or destroyed) in a chemical reaction, we must always balance the chemical equation Stoichiometry Balanced: CO(g) + 2H2(g) → CH3OH(l) • It is important to recognize that the coefficients in a balanced equation give the relative number of molecules • That is, we could multiply this balanced equation by any number and still have a balanced equation • For example, we could multiply by 12, • 12 [ CO(g) + 2H2(g) → CH3OH(l) ] to obtain • 12 CO(g) + 24H2(g) → 12CH3OH(l) Stoichiometry • 12 CO(g) + 24H2(g) → 12CH3OH(l) • This is still a balanced equation • Because 12 represents a dozen, we could even describe the reaction in terms of dozens: 1 dozen CO(g)+ 2 dozen H2(g) →1 dozen CH3OH(l) • We could also multiply the original equation by a very large number, such as 6.022 x 1023 Stoichiometry • 6.022 x 1023 [ CO(g) + 2H2(g) → CH3OH(l) ] • Which leads to the equation below: 6.022 x 1023 CO(g) + 2(6.022 x 1023 )H2(g) → 6.022 x 1023 CH3OH(l) • We also know this number 6.022 x 1023 is equal to What????? • Our equation can be written in terms of moles: 1 mol CO(g) + 2 mol H2(g) → 1 mol CH3OH(l) Mole – Mole Relationships • Now that we have discussed the meaning of a balanced chemical reaction in terms of moles of reactants and products, we can use an equation to predict the moles of products that a given number of moles of reactants will yield • For example, consider the reaction of the decomposition of water: • 2H2O(l) → 2H2(g) + O2(g) • The equation tells us that 2 mol of H2O yields 2 mol of H2 and 1 mol of O2 Mole – Mole Relationships • Now suppose we decompose 5.8 mol of water • What number of moles of products are formed in this process? • We could answer this question by re-balancing the equation as follows: • First, we divide all coefficients of the balanced equation by 2, • 2H2O(l) → 2H2(g) + O2(g) This gives: • H2O(l) → H2(g) + ½ O2(g) Mole – Mole Relationships • Now because we have 5.8 mol of water, we multiply this equation by 5.8 • 5.8 [H2O(l) → H2(g) + ½ O2(g) ] which gives, • 5.8H2O(l) → 5.8H2(g) + 5.8(½) O2(g) • 5.8H2O(l) → 5.8H2(g) + 2.9 O2(g) • So, 5.8 mol of water yields 2.9 mol of of Oxygen and 5.8 mol of hydrogen Mole Ratio • Take a look at the equation below: 2H2O(l) → 2H2(g) + 1O2(g) • The above equation tells us that 2 moles of H2O will produce 2 moles of H2 and 1 mol of O2 2 moles H2O or __2 moles H2___ • We can write this as a ratio: 2 moles H2 2 moles H2O • Or, we can use this ratio: 2 moles H2O 1 moles O2 or 1 moles O2___ 2 moles H2O __ • This ratio is important because it allows us to go from moles of reactants to moles of products • The mole ratio is conversion factor that allows us to go from moles of reactants to moles of products moles of A on the reactant side moles B on the product What is the mole ratio? conversion factor that takes • A __________ us from moles of ________ reactants to moles of ________ products moles of A on the reactant side moles B on the product Answer Bank Reactants conversion moles Quantities Reactions products Practice: First balance the equations below and determine the mole ratios • 3NO2(g) + H2O (l) → 2 HNO3(aq) + NO (g) • Determine the mole ratio for NO2 and HNO3 3 moles of NO2 2 moles of HNO3 or 2 moles of HNO3 3 moles of NO2 • C6H6 (g) + 3 H2 (g) → C6H12(g) • Determine the mole ratio for H2 and C6H12 3 moles of H2 1 moles of C6H12 or 1 moles of C6H12 3 moles of H2 Practice: • Using the equation below, what number of moles of O2 will be produced by the decomposition of 6.4 mol of water • 2H2O(l) → 2H2(g) + O2(g) 6.4 mol H2O 1 mol O2 = 3.2 mol of O2 2 mol H2O Practice: • Using the equation below, calculate the number of moles of NH3 that can be made from 1.3 mol H2 reacting with excess N2 • N2(g)+ 3H2(g) → 2NH3(g) 1.3 mol H2 2 mol NH3 3 mol H2 = 0.867 mol of NH3 Practice: • How many moles of N2 are needed to produce 8.5 moles of NH3? • N2(g)+ 3H2(g) → 2NH3(g) 8.5 mol NH3 1 mol N2 2 mol NH3 = 4.25 mol of N2 Practice: • Using the equation below, calculate the number of grams of NH3 that can be made from 3.9 mol H2 reacting with excess N2 • N2(g)+ 3H2(g) → 2NH3(g) 3.9 mol H2 2 mol NH3 3 mol H2 17.0 g of NH3 1 mol NH3 = 44.2 grams of NH3 Summarize: • Stoichiometry is…….. • The ____ ____is a conversion factor that takes us from moles of _____ to moles of ____ • When performing stoichiometry problems always make sure that your _____ cancel • Review: A solid that dissolves in water is ______ & a solid that does not dissolve in water is _______ Unit: Stoichiometry Topic: Mass Calculations Objectives: Day 2 of 4 • To learn how to perform mass calculations (convert grams of a product into grams of a reactant) Quickwrite Answer one of the questions below 1-2 sentences: • In chemistry, we count by weighing moles; We can’t measure moles in a lab, BUT what unit of measurement can we use in the lab to count atoms or molecules??? • Review: Using the equation below, 2 moles of H2O will produce how many moles of O2 and H2? 2H2O(l) → 2H2(g) + 1O2(g) Mass Calculations • We just saw how to use balanced equations for a reaction to calculate the numbers of moles • Remember, moles represent numbers of molecules and we cannot count molecules directly • In chemistry, we count by weighing • When we weigh we use the gram, therefore we need to learn how to convert moles to mass Mass Calculations • Let’s consider an unbalanced combustion reaction in which propane reacts with oxygen to produce carbon dioxide and water • C3H8(g) + O2(g) → CO2(g) + H2O(g) • What mass of oxygen will be required to react exactly with 44.1 grams of propane? Mass Calculations • First, we need to balance the equation: • C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) • Let’s summarize what we know and what we want to find • What we know: – The balanced equation for the reaction – The mass or amount of propane availible(44.1g) • What we want to calculate: – The mass of oxygen required to react exactly with all the propane Mass Calculations 44.1 g propane requires ? Grams of O2 C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) Our overall plan of attack is as follows: 1. We are given the number of grams of propane, so we must convert to moles of propane (C3H8), because the balanced equation deals in moles not grams 2. Next, we can use the coefficients in the balanced equation to determine the moles of oxygen(O2) required 3. Finally, we will use the molar mass of O2 to calculate grams of oxygen Our Plan of Attack! We are given grams of propane We have to convert grams of propane into moles 44.1 g propane ? moles of propane Use mole ratio 2 convert moles of propane into moles of O2 We have to convert moles into grams of O2 ? moles of O2 ? Grams of O2 44.1 g C3H8 requires ?? g O2 C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) ••• Next, wegiven can use the the We are the number of grams of in propane, Finally, we will use thecoefficients molar mass of so to wecalculate must convert to moles of propane (C3H8), balanced equation to determine the moles O grams of oxygen 2 the balanced equation deals in moles ofbecause oxygen(O 2) required not grams 44.1 g C3H8 1 mol C3H8 5 mol O2 32.0 g O2 44.09 g C3H8 1 mol C3H8 1 mol O2 = 160 g of O2 Practice: 2Al(s) + 3I2(g) → 2AlI3(s) • Consider the above reaction • Calculate how many grams of the product aluminum Iodide (AlI3)would be produced by the complete reaction of 35.0 grams of Aluminum (Al) Our Plan of Attack! We are given grams of Al We have to convert grams of Al into moles of Al Use mole ratio 2 convert moles of Al into moles of AlI3 35.0 g of Al ? moles of Al ? moles of AlI3 2Al(s) + 3I2(g) → 2AlI3(s) We have to convert moles into grams of AlI3 ? Grams of AlI3 2Al(s) + 3I2(g) → 2AlI3(s) • Back to the problem!!!! • Calculate how many grams of the product aluminum Iodide (AlI3)would be produced by the complete reaction of 35.0 grams of Aluminum (Al) 35.0 g Al 1 mol Al 26.98 g Al 2 mol AlI3 407.68 g AlI3 2 mol Al 1 mol AlI3 = 528.87 g of AlI3 Practice: Br2(l) + 2NaCl(aq) → 2NaBr(aq) + Cl2(g) • Consider the above balanced reaction • Calculate how many grams of the product chlorine(Cl2) be produced by the complete reaction of 5.0 grams of sodium chloride (NaCl) 5.0 g NaCl 1 mol NaCl 1 mol Cl2 70.9 g Cl2 58.43 g NaCl 2 mol NaCl 1 mol Cl2 = 3.03 g of Cl2 Practice: • • How many grams of chlorine gas are needed to produce 10.0 g of sodium chloride? Remember to balance first!!!!! Cl2 + 2NaI 2 NaCl + I2 10.0 g NaCl 1 mol NaCl 1 mol Cl2 70.9 g Cl2 = 6.06 g Cl2 58.43 g NaCl 2 mol NaCl 1 mol Cl2 Summarize: (fill in the blank) You are given ____ of reactant Grams of reactant Use Molar mass Of reactant to Get moles of ______ ___ of reactant Use mole _____ To get from Moles of reactant To moles of product Moles of product Answer Bank Product Ratio Mass moles reactant grams Use Molar ___ Of product to Get grams of product Grams of_____ Unit: Stoichiometry Topic: Limiting Reactant Objectives: Day 3 of 4 • To learn what the limiting reactant is in a chemical reaction • To learn how to calculate the limiting reactant using moles of reactants Quickwrite Answer one of the questions below 1-2 sentences: • A sandwich recipe requires 2 pieces of bread, 3 slices of meat and 1 slice of cheese; you go into the kitchen and realize that you have 2 pieces of bread, 1 slice of cheese, and NO MEAT; what limited affected your ability to make your sandwich??? Limiting Reactants • Earlier, we discussed making sandwiches • Recall, that the sandwich making process could be described as follows: 2 pieces of bread + 3 slices of meat + 1 slice of cheese →1 sandwich • In this equation, all the products are used up, nothing was left over • Now assume you came to work one day and found the following quantities of ingredients Limiting Reactants • Now assume you came to work one day and found the following quantities of ingredients • 20 slices of bread • 24 slices of meat • 12 slices of cheese • How many sandwiches can you make? • What will be left over Limiting Reactants • To solve this problem, let’s see how many sandwiches we can make with each ingredient:How many sandwiches • Bread: • Meat: can you of make? answer 1 sandwich 20 slices bread The is 8! Once you run out of meat, 2 slices of bread you must stop making sandwiches. The meat is the limiting ingredient! 24 slices of meat =10 sandwiches 1 sandwich 3 slices of meat = 8 sandwiches • Cheese: 12 slices of cheese 1 sandwich 1 slice of cheese =12 sandwiches Limiting Reactants • What do you have left over? • Making 8 sandwiches requires 16 pieces of bread • You started with 20 slices, so you have 4 slices of bread left over • You also used 8 pieces of cheese for the 8 sandwiches, so you have 4 pieces of cheese left over • In this example, the meat was the limiting reactant Limiting Reactants • When molecules react with each other to form products, considerations very similar to those making sandwiches arise • Consider the reaction that occurs when Hydrogen Gas reacts with oxygen gas to form water H2(gas) + O2(gas) H2O(liquid) Limiting Reactants • The reaction occurs between 10 H2 molecules and 7 O2 molecules • Remember, each O2 molecule requires 2 H2 molecules 2H2(gas) + O2(gas) 2H2O(liquid) Limiting Reactants • After the reaction, 10 water molecules formed and 2 O2 molecules are left over • That is, the H2 molecules are used up before the water molecules are consumed • We have excess (extra) O2 and H2 is the limiting reactant because the reaction runs out of H2 first 2H2(gas) + O2(gas) 2H2O(liquid) What is the Limiting Reactant? consumed reactant that is completely _________ • The _______ when a reaction is run to completion • The reactant that is not completely consumed excess is in _____ Answer Bank Moles consumed product Limiting reactant excess Practice: N2 + 3 H2 2 NH3 You have 2 moles of N2 and 7 moles of H2 Given: 2 mole of N2 7 mol of H2 Which is the limiting reactant? 7 mol H 2 2.3 3mol H 2 2mol N 2 2 1mol N 2 N2 is the limiting reactant because it has the lowest ratio!!!!!!!!!!!!!! Practice: N2 + 3 H2 2 NH3 You have 3 moles of N2 and 6 moles of H2 Given: 3 mole of N2 6 mol of H2 Which is the limiting reactant? 6 moles of H2 3 moles of H2 3 moles of N2 1 moles of N2 H2 is the limiting reactant because it has the lowest ratio!!!!!!!!!!!!!! Practice: This reaction is25 different the othersreacted we have done • Suppose gramsfrom of nitrogen So far 5in grams that weof arehydrogen mixing specified amounts with gas are mixedof Twoand reactants together. To know how much product react to form ammonia. Calculate the Forms we must we must determine which reactant is mass of ammonia produced when this consumed first. In other words, we must determine reaction is The run limiting to completion reactant • N2(g) + 3H2(g) → 2NH3(g) Practice: • Suppose 25 grams of nitrogen reacted with 5 grams of hydrogen gas are mixed and react to form ammonia. Calculate the mass of ammonia produced when this reaction is run to completion • N2(g) + 3H2(g) → 2NH3(g) 25.0 g N2 1 mol N2 28.0 g N2 5.0 g H2 1 mol H2 2.016 g H2 = .892 moles of N2 = 2.48 moles of H2 Now we must determine which reactant Practice: is the limiting reactant. We have 0.892 moles of nitrogen • Suppose 25 grams ofmany nitrogen reacted with 5 Let’s determine how moles of hydrogen grams of hydrogen gaswith arethis mixed andnitrogen. react to form Are required to react much ammonia.1 mol Calculate the mass of with ammonia Because of nitrogen reacts 3 mol of produced when this reaction is run to completion Hydrogen, the number of moles of hydrogen we Need completely with 0.892 mol • N2(g) + 3Hto2(g)react → 2NH 3(g) of nitrogen is calculated as follows: 0.892 mol N2 3 mol H2 1 mol N2 = 2.68 moles of H2 Is nitrogen or hydrogen the limiting reactant? The answer comes from the comparison: 2.48 moles of H2 available < 2.68 moles of H2 required This means that the hydrogen will be consumed first before the nitrogen Runs out, so hydrogen is the limiting reactant Reflect: Practice: We see that 0.892 mol of nitrogen require 2.68 mol Of hydrogen to react completely. However, only 2.48 Mol of hydrogen available. This means that the hydrogen • Suppose 25are grams of nitrogen reacted with 5 willgrams be consumed before the runs out, Hydrogen of hydrogen gasnitrogen are mixed andso react to form Is the limiting reactant ammonia. Calculate the mass of ammonia producedIfwhen this reaction isthen run the to completion the nitrogen is excess, • N2(g) +hydrogen 3H2(g) → will 2NH run out 3(g)first. Again we find that the hydrogen limits the amount of ammonia Formed 25.0 g N2 2 mol NH 17.0 g NH 2.48 mol Hthe 2 Because moles of3 hydrogen 3are limiting, we must use = the 28.1moles g of NHof3 ammonia Our quantity of Hydrogen to determine 3 mol H2 1 mol NH3 that can form Summarize: • When determining the limiting reactant we first convert mass into_____ • The smaller molar value of the reactants is the _____ reactant and the larger molar value of the reactants is in excess • The limiting reactant will determine how much ______ will form Answer Bank Moles consumed product Limiting reactant excess Unit: Stoichiometry Topic: Theoretical & Percent Yield Objectives: Day 4 of 4 • To understand what theoretical yield is • To learn how to calculate percent yield using the theoretical yield Quickwrite Answer one of the questions below 1-2 sentences: • Chemists who work for companies like to know how efficient a reaction is, in other word, they would like to know how much product forms. Why do you think they would want to know this??? Percent Yield • If you recall, the amount of product formed is determined by the limiting reactant • The amount of product calculated using stoichiometry is called the theoretical yield • The theoretical yield is the amount of product predicted from the amounts of reactants used up • Think of it as the maximum amount of product that could be produced from 100% of the reactants being used up • The theoretical yield is rarely if ever actually obtained What is the theoretical yield? calculated maximum • The ________ amount of product that can be formed when the limiting reactant is completely used up or ________ consumed • It is calculated using ___________ stoichiometry Answer Bank actual Consumed percentage stoichiometry Calculated theoretical Percent Yield • Why is the theoretical yield never reached? • One reason for this is the presence of side reactions that consume some of the reactants • The actual yield of product, which is the amount of product actually obtained, is compared to the theoretical yield • This comparison, called the percent yield, is expressed as a percent • The percent yield is calculated by dividing the actual yield / by the theoretical yield • Percent yield = Actual yield x 100% Theoretical yield What is the percent yield? actual (experimental) yield of a • The ______ percentage over of the product given as a _________ __________ theoretical yield • Percent yield = Actual yield x 100% Theoretical yield Answer Bank actual Consumed percentage stoichiometry Calculated theoretical Practice: • Consider the balanced reaction below: • 2H2(g) + CO (g) → CH3OH(l) • Suppose 68.5 grams of CO is reacted with 8.6 grams of H2 • Calculate the theoretical yield of methanol CH3OH(l) • Your experiment actually produces 35.7 grams of methanol CH3OH(l) • What is the percent yield of methanol CH3OH(l) ? Practice: • Consider the balanced reaction below: Step 1: Calculate the moles of reactants • 2H2(g) + CO (g) → CH3OH(l) • Suppose 68.5 grams of CO is reacted with 8.6 grams of H2 • Calculate the theoretical yield of methanol CH3OH(l) 68.5 g CO 1 mol CO 28.01 g CO 8.60 g H2 1 mol H2 2.016 g H2 = 2.45 moles of CO = 4.27 moles of H2 Practice: Step 2: Now we determine reactant is reaction limiting using • Considerwhich the balanced below: The mole ratio of between CO and H2 • 2H2(g) + CO (g) → CH3OH(l) • Suppose 68.5 grams of CO is reacted with 8.6 grams of H2 • Calculate the theoretical yield of methanol CH3OH(l) 2.45 mol CO 2 mol H2 1 mol CO = 4.9 moles of H2 Is CO or H2 the limiting reactant? The answer comes from the comparison: 4.27 moles of H2 present < 4.9 moles of H2 needed to react with all the CO This means that the hydrogen will be consumed first before the CO Runs out, so hydrogen is the limiting reactant Practice: Step 3: We must use the amount of H and • Consider the balanced reaction below: the mole ratio between CH3OH and H2 to determine • 2H + CO the 2(g) maximum amount methanol that can be produced (g) →of CH 3OH(l) • Suppose 68.5 grams of CO is reacted with 8.6 grams of H2 • Calculate the theoretical yield of methanol CH3OH(l) 4.27 mol H2 1 mol CH3OH(l) 2 mol H2 2 = 2.14 moles of CH3OH 2.14 represents the theoretical yield Which if you recall, is the maximum amount of a given product that can be formed when the limiting reactant is completely used up or consumed Practice: • Consider the balanced reaction below: Step 4: Convert moles to grams • 2H2(g) + CO (g) → CH3OH(l) • Suppose 68.5 grams of CO is reacted with 8.6 grams of H2 • Calculate the theoretical yield of methanol CH3OH(l) 2.14 mol CH3OH 32.04 g CH3OH 1 mol CH3OH = 68.6 grams of CH3OH Practice: • Consider the balanced below: the Step 5: Calculate percentreaction yield by dividing (35.7 g) by the theoretical yield (68.6 g) •actual 2H2(g)yield + CO → CH OH (g) 3 (l) • Suppose 68.5 grams of CO is reacted with 8.6 grams of H2 • Calculate the theoretical yield of methanol CH3OH(l) • Your experiment actually produces 35.7 grams of methanol CH3OH(l) • Percent yield = 35.7 g CH3OH x100 = 52% 68.6 g CH3OH Summarize: • The _____ ______ is the calculated maximum amount of product that can be formed when the limiting reactant is completely used up or consumed • The _____ ______ is the actual (experimental) yield of a product given as a percentage over of the theoretical yield • % Yield = ___????____ x 100% ????