# Chemistry Stoichiometry Lesson 8 Lesson Plan David V. Fansler

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1 Chemistry Stoichiometry Lesson 8 Lesson Plan David V. Fansler The Arithmetic of Equations Objectives: Interpret balanced chemical equations in terms of interacting moles, representative particles, masses, and gas volume at STP - The Story So Far o We have learned chemical symbols, charges, how to put together a chemical formula, how to write a chemical reaction and lastly how to balance a chemical reaction o Now we enter the world of Stoichiometry determining how much of an element or compound is needed to make a certain amount of product, or visa versa - Stoichiometry o The calculations of quantities in chemical reactions. o We use moles, atoms, molecules, representative particles, grams and liters o Take for example the creation of ammonia; N( g) + H( g) NH( g) So what can we get from this balanced equation Particles We can see that 1 molecule of nitrogen reacts with molecules of hydrogen to form molecules of ammonia based on this there is a 1:: ratio. Since dealing with just a few molecules is impossible, we could scale it up to be 1 Avogadro s number of nitrogen molecules reacting with Avogadro s number of hydrogen molecules to form Avogadro s number molecules of ammonia. Moles Since we know that 1 mole is equal to 1 Avogadro s number, we can see that it takes 1 mole of nitrogen and moles of hydrogen to form moles of ammonia a ratio of 1:: So we see that in a balanced equation, the coefficients represent the number of moles of each element/compound in the reactant and product this is very important information. Using this information you can calculate the amount of reactants and products. For the reaction forming NH, does the number of moles of reactant equal the number of moles of product? NO Mass remember the law of conservation of mass? It still applies here as well. The sum of the mass of all the reactants must be equal to the mass of the product. So in our example of forming ammonia, 1 mole of nitrogen has a mass of 8.0 g, and moles of hydrogen has a mass David V. Fansler Beddingfield High School Page 1

2 - of 6.0 g for a total of 4.0 g of product. Since ammonia is NH, one mole would have a mass of 17.0g, therefore the two moles produced would have a mass of 4.0g, which does equal the mass of the reactants. Mass is indeed conserved Volume As we recently learned, it is easier to express gases in terms of volume. We used STP to determine a common temperature and pressure for measuring volumes. As well we learned that 1 mole of a gas is.4 liters. So in the production of ammonia, we see that 1 mole of nitrogen gas (.4 l) reacts with moles of hydrogen gas (67. l) to form moles of ammonia (44.8 l) gas The below table shows how mass and atoms are conserved in every reaction, but that molecules, formula units, moles and volumes of gases will not necessarily be conserved N (g) + H (g) NH (g) + atoms N + 6 atoms H atoms N and 6 atoms H 1 molecule N + molecules H molecules NH 10 molecules N + 0 molecules H 0 molecules NH 1 x (6.0 x 10 ) + x (6.0 x 10 ) x (6.0 x 10 ) (molecules N ) (molecules H ) (molecules NH ) 1 mol N + mol H mol NH 8.0 g N g H 4.0 g NH 4.0 g of reactants 4.0 g of product Assume STP.4 L +.4 L.4 L.4 L.4 L.4 L N H H H NH NH.4 L N L H 44.8 L NH David V. Fansler Beddingfield High School Page

3 Sample Problems Hydrogen sulfide, a foul smelling gas (the smell of rotten eggs) is often found around volcano vents. From the balanced equation below, give the interaction of the three relative quantities: a. number of representative particles b. number of moles c. masses of reactants and products HSg ( ) + O( g) SO( g) + H0( g) a. molecules of H S react with molecules of O to produce molecules of SO and molecules of H O b. moles of H S react with moles of O to produce moles of SO and molecules of H O c. multiply the number of moles times the molar mass for each reactant and product 4.1g.0g 64.1g 18.0g mol x + mol x mol x + mol x mol mol mol mol 68.gHS gO 18.gSO + 6.0gHO 164.g = 164.g Balance the equation for the combustion of acetylene and interpret the equation in terms of relative numbers of moles, volumes of gas at STP, and masses of reactants and products. CH ( g) + O( g) CO( g) + HOl ( ) CH ( g) + 5 O( g) 4 CO( g) + HOl ( ) o mol C H + 5 mol O + 4 mol CO + mol H O o 44.8 L C H L O 89.6 L CO L of H O 6.0g.0g 44.0g 18.0g mol x + 5mol x 4mol x + mol x mol mol mol mol o 5.0gCH gO 176.0gCO + 6.0gHO 1.0g rea tan ts 1.0g products -Chemical Calculations Objectives: Construct mole ratios from balanced chemical equations and apply these ratios in mole-mole stoichiometric calculations; Calculate stoichiometric quantities from balanced chemical equations using units of moles, mass, representative particles and volumes of gases at STP. - Mole-Mole Calculations o With a balanced equation, we know the relationship of how many moles of reactants are needed to make so many moles of product. So if we know the number of moles of one substance, we can determine the number of moles of all other substances in the reaction o Look at the production of ammonia again David V. Fansler Beddingfield High School Page

4 + N ( g) H ( g) NH ( g) As noted earlier, the most important interpretation of this equation is that 1 mol of N reacts with moles of H to from moles of NH it is with this ratio knowledge that we can relate the number of moles of reactant to product The coefficients from the balanced equation are used to write conversions called mole ratios Mole ratios are used to calculate the number of moles of product from a given number of reactants, or to calculate the number of moles of reactant from a given number of moles of product. For the production of ammonia we have a ratio of 1N :O :NH, so we can set up the following ratios 1 moln 1, moln, molh mol H mol NH molnh Or the reverse of these as mol H, mol NH mol NH, 1molN 1molN molh o So how do we use Mole-Mole Ratios? Sample Problems How many moles of ammonia are produced when 0.60 mol of nitrogen reacts with hydrogen? We set up a conversion using one of the mole ratios where we use 0.60 mol N where the N will cancel out and leave us with NH mol NH 0.60mol N x 1.mol NH 1mol N = Given the equation for the formation of aluminum oxide: 4 Al( s) + O( g) AlO( s) a. Write the 6 mole ratios that can be derived b. How many moles of aluminum are needed to form.7 mol Al O? a. 4mol Al 4mol Al mol O mol O,,,, mol Al O mol Al O, mol O mol AlO mol AlO 4mol Al 4mol Al mol O 4mol Al b..7mol AlO x 7.4mol Al mol Al O = According to the above equation: a. How many moles of oxygen are required to react completely with 14.8 mol Al? b. How many moles of Al O are formed when 0.78 mol O reacts with aluminum? David V. Fansler Beddingfield High School Page 4

5 mol O a. 14.8mol Al x 11.1mol O 4mol Al = mol AlO b mol O x.5mol AlO mol O = - Mass-Mass Calculations o Scales tend to read in mass, not moles, since that would involve counting atoms/molecules which we cannot do! So we need to be able to do the same mole-mole calculations with grams, something we can measure o The method is one where we convert the mass to moles using molar mass, use the mole ratio, and then convert back to mass using molar mass Sample Problem Calculate the number of grams of NH produced by the reaction with 5.40 g of hydrogen with an excess of nitrogen. N( g) + H ( g) NH( g) balanced equation 1mol H 5.40g H x =.7mol H convert g H to mole H.0gH mol NH.7mol H x = 1.8mol NH find mole of NH mol H 17.0gNH 1.8mol NH x = 0.6 NH convert mol NH to g NH 1mol NH Acetylene gas (C H ) is produced by adding water to calcium carbonate (CaC ). How many grams of acetylene are produced by adding water to 5.00 g of CaC. CaC() s + HO() l CH ( g) + Ca( OH )( aq) CaC() s + H O() l CH ( g) + Ca( OH )( aq) balanced 1mol CaC 5.00 g CaC x =.078mol CaC 64.1gCaC 1mol C H.078 mol CaC x =.078mol C H 1mol CaC 6.0gCH.078mol C H x =.0g C H 1mol CH Using the same equation, how many moles of CaC are needed to react completely with 49.0g H O? David V. Fansler Beddingfield High School Page 5

6 CaC() s + H O() l CH ( g) + Ca( OH )( aq) balanced 1mol H O 49.0gHOx =.7molHO 18.0gHO 1mol CaC.7mol H O x = 1.6mol CaC mol H O o Now that we know how to do mass-mass problems, could we do volumevolume, volume-mass, mass-representative problems? The answer is yes! It is just a matter of conversion. Representative particles to moles Mass to moles Volume to moles Returning to moles by Moles to representative particles Moles to mass Moles to volume Sample Problems How many molecules of oxygen are produced when a sample of 9. g of water is decomposed by electrolysis according to this balanced equation? HOl ( ) H( g) + O( g) 1mol H O 9.gHOx = 1.6 molho 18.0gHO 1mol O 1.6 mol H O x =.810 mol O mol H O 6.0 x 10 molecules O = 1 mol O.810 mol O x 4.88 x 10 molecules O How many molecules of oxygen are produced by the decomposition of 6.54g of potassium chlorate (KClO )? KClO KCl + O KClO KCl + O 1mol KClO 6.54 KClO x =.05mol KClO 1.6gKClO mol O.05 mol KClO x =.0800 mol O mol KClO 6.0 x 10 molecules O.0800 mol O x = 4.8 x 10 molecules O 1mol O In making nitric acid, the last reaction is nitrogen oxide and water. How many grams of nitrogen must react with water to produce 5.00 x 10 molecules of nitrogen monoxide? David V. Fansler Beddingfield High School Page 6

7 NO ( g) + H O( l) HNO ( aq) + NO( g) 1mol NO 5.00 x 10 molecules NO x = 8.1 x 10 mol NO 6.0 x 10 molecules NO mol NO x mol NO x mol NO 1mol NO = gNO.49mol NO x = 11.5g NO 1mol NO Assuming STP, how many liters of oxygen are needed to produce 19.8 L SO according to this balanced equation? SO( g) + O( g) SO( g) 1mol SO 19.8 LSO x =.884molSO.4LSO 1mol O.884 mol SO x =.44mol O mol SO.4LO.44mol O x = 9.90 L O 1mol O Using the equation for the combustion of carbon dioxide, how many liters of oxygen are required to completely burn.86l of carbon monoxide? CO + O CO 1mol CO.86 LCOx =.17molCO.4LCO 1mol O.17 mol CO x =.0860 mol O mol CO.4LO.0860 mol O x = 1.9L O 1mol O Limiting Reagent and Percent Yield Objectives: Identify and use the limiting reagent in a reaction to calculate the maximum amount of product(s) produced and the amount of excess reagent; Calculate the theoretical yield, actual yield, or percent yield given appropriate information. - Limiting Reagent o In working on mole-mole and mass-mass problems, we have made statements similar to When 6.4 g of N react with an excess of H, how much NH is produced? The key to this statement is with an excess of H. The answer would be 1moleN mol NH 17.0g NH 6.4gN x x x 7.8g NH 8.0gN 1molN 1molNH = David V. Fansler Beddingfield High School Page 7

8 o Suppose, instead of an excess of H, there was only 1.0g available then how much NH could we have produced? Would 1.0 g H be enough to react completely with 6.4g N? o To answer this problem, we choose one of the reactants and solve for the amount of the other that would be needed for the given amount. 1moleN mol H.0g H o 6.4gN x x x 1.4g H 8.0gN 1mol N 1mol H = o So we see that to react with 6.4g of N we would need 1.4g H and we only have 1.0g H. This means that H is the limiting reagent o On the other hand, since H is the limiting reagent, since we have more than enough N for 1.0g H, then N is the excess reagent. Sample Problem Sodium chloride can be prepared by the reaction of sodium metal with chlorine gas. Suppose that 6.70 mol of Na reacts with.0 mol Cl. a. What is the limiting reagent b. How many moles of NaCl are produced? Na( s) + Cl( g) NaCl( s) 1mol Cl 6.70mol Na x =.5mol Cl Cl is the lim iting reagent mol Na mol NaCl.0mol Cl x = 6.40mol NaCl 1mol Cl - Calculating the Percent Yield o When chemical reactions take place, they are almost never 100% complete. A reaction may not go to 100% due to not all the reactants becoming involved, impurities in the reactants, loss of product due to filtering, or just not getting it all out of the vessel. o The predicted amount of product which we have been doing with our Stoichiometry problems has been for 100% yield, or the theoretical yield o The actual yield is how much product can be collected and measured. o The percent yield is a ratio given by: Actual Yield Percent Yield = x 100% Theoretical Yield Sample Problem Calcium carbonate is decomposed by heating, as shown in the following equation: CaCO() s CaO() s + CO( g) a. What is the theoretical yield of this reaction if 4.8g CaCO is heated to five 1.1g CaO? b. What is the percent yield? 1mol CaCO 1mol CaO 56.1g CaO 4.8gCaCO x x x = 1.9gCaO 100.1g CaCO 1mol CaCO 1mol CaO 1.1gCaO Percent Yield = x 100% = 94.% 1.9gCaO - David V. Fansler Beddingfield High School Page 8

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