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Stoichiometry

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9.1 Calculating Quantities in Reactions Determine mole ratios from a balanced chemical equation Explain why mole ratios are central to solving stoichiometry problems Solve stoichiometry problems involving: – Mass using molar mass – Volume using denisty – Particles using Avogadro’s number

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Ham Sandwich How many sandwiches could you make from 24 slices of bread? How many slices of lettuce would you need? tomato? ham? cheese? This process models the calculations in this chapter.

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Equations are like recipes. 2C 8 H 18 + 25O 2 → 16CO 2 + 18H 2 O Coefficients can be read as ratios of particles or of moles. If 2 mols of C 8 H 18 reacts completely how many moles of CO 2 would be produced?

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Stoichoimetry = the proportional relationship between 2 or more substances during a chemical reaction quantitative analysis of the outcomes of a reaction Predict amount of product you could make from starting amounts of reactan t

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The Mole Ratio is the Key In stoichiometry problems, the unit that bridges the gap between one substance and another is the mole. You can use the coefficients in conversion factors called mole ratios

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Sample Problem A N 2 + 3H 2 → 2NH 3 3 mols of hydrogen are needed to prepare 2 moles of ammonia. 3 mol H 2 = 2 mol NH 3 PROBLEM: How many moles of hydrogen are needed to prepare 312 moles of ammonia?

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312 mol NH 3 x 3 mol H 2 = ? mol H 2 2 mol NH 3 = 468 mol H 2 needed

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Practice Prob. P. 304 1.Calculate the amounts requested if 1.34 mol H 2 O 2 completely react according to the following equation 2H 2 O 2 → 2H 2 O + O 2 a.mols of oxygen formed Given:1.34 mol H 2 O 2 reacts 2 mol H 2 O 2 = 1 mol O 2 1.34 mol H 2 O 2 x 1 mol O 2 = 0.670 mol O 2 produced 2 mol H 2 O 2

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Practice Prob. P. 304 1.Calculate the amounts requested if 1.34 mol H 2 O 2 completely react according to the following equation 2H 2 O 2 → 2H 2 O + O 2 b. mols of water formed Given:1.34 mol H 2 O 2 reacts 2 mol H 2 O 2 = 2 mol H 2 O 1.34 mol H 2 O 2 x 2 mol H 2 O = 1.34 mol H 2 O produced 2 mol H 2 O 2

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2. Calculate the amounts requested if 3.30 mol Fe 2 O 3 completely react according to the following equation Fe 2 O 3 + 2Al → 2Fe + Al 2 O 3 Thermite Reaction Thermite Reaction a.mols of aluminum needed Given:3.30 mol Fe 2 O 3 reacts 1 mol Fe 2 O 3 = 2 mol Al needed 3.30 mol Fe 2 O 3 x 2 mol Al = 6.60 mol Al needed 1 mol Fe 2 O 3

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2. Calculate the amounts requested if 3.30 mol Fe 2 O 3 completely react according to the following equation Fe 2 O 3 + 2Al → 2Fe + Al 2 O 3 b. mols of iron formed Given:3.30 mol Fe 2 O 3 reacts 1 mol Fe 2 O 3 = 2 mol Fe formed 3.30 mol Fe 2 O 3 x 2 mol Fe = 6.60 mol Fe formed 1 mol Fe 2 O 3

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2. Calculate the amounts requested if 3.30 mol Fe 2 O 3 completely react according to the following equation Fe 2 O 3 + 2Al → 2Fe + Al 2 O 3 c. mols of aluminum oxide formed Given:3.30 mol Fe 2 O 3 reacts 1 mol Fe 2 O 3 = 1 mol Al 2 O 3 formed 3.30 mol Fe 2 O 3 x 1 mol Al 2 O 3 = 3.30 mol Al 2 O 3 formed 1 mol Fe 2 O 3

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Application/Homework Worksheet “Stoichiometry: Mole-Mole Problems” – front only

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Stoichiometry: Mass-Mass Problems You must convert to moles using molar mass of known Then use mole ratio to find moles unknown Convert back to mass using molar mass of unknown

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Solving Mass-Mass Problems Molar mass Mass Known Molar Ratio Mol Known Molar Mass Mol Unknown Mass Unknown 1 mol ___ grams Periodic Table Mol unknown Mol known Balanced Chemical Equation ___ grams 1 mol Periodic Table

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Sample Problem B p. 307 What mass of NH 3 can be made from 1221 g H 2 and excess N 2 ? N 2 + 3H 2 → 2 NH 3 1221 g H 2 X 1 mol H 2 x 2 mol NH 3 x 17.04g NH 3 2.02 g 3 mol H 2 1 mol NH 3 GRAMS MOLAR MOLAR MOLAR KNOWN MASS RATIO MASS KNOWN UNKNOWN = 6867 g NH 3 made

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Practice p. 307#1-4 Use the equation below to answer #1-4 Fe 2 O 3 + 2Al → 2Fe + Al 2 O 3 1. How many grams Al needed to completely react with 135 grams Fe 2 O 3 135 g Fe 2 O 3 X 1 mol Fe 2 O 3 x 2 mol Al x 26.98 g Al 159.7 g Fe 2 O 3 1 mol Fe 2 O 3 1 mol Al GRAMS MOLAR MOLAR MOLAR KNOWN MASS RATIO MASS KNOWN UNKNOWN = 45.6 g Al needed

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Practice p. 307#1-4 Use the equation below to answer #1-4 Fe 2 O 3 + 2Al → 2Fe + Al 2 O 3 2. How many grams Al 2 O 3 can form when 23.6g Al react with excess Fe 2 O 3 ? 23.6 g Al X 1 mol Al x 1 mol Al 2 O 3 x 101.96 g Al 2 O 3 26.98 g Al 2 mol Al 1 mol Al 2 O 3 = 44.6 g Al 2 O 3 formed

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Practice p. 307#1-4 Use the equation below to answer #1-4 Fe 2 O 3 + 2Al → 2Fe + Al 2 O 3 3. How many grams of Fe 2 O 3 react with excess Al to make 475 g Fe? 475 g Fe X 1 mol Fe x 1 mol Fe 2 O 3 x 159.7 g Fe 2 O 3 55.85 g Fe 2 mol Fe 1 mol Fe 2 O 3 = 679 g Fe 2 O 3 react

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Practice p. 307#1-4 Use the equation below to answer #1-4 Fe 2 O 3 + 2Al → 2Fe + Al 2 O 3 4. How many grams of Fe will form when 97.6 g Al 2 O 3 form? 97.6 g Al 2 O 3 X 1 mol Al 2 O 3 x 2 mol Fe x 55.85 g Fe 101.96 g Al 2 O 3 1 mol Al 2 O 3 1 mol Fe = 107 g Fe formed

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Homework Worksheet “Stoichiometry: Mass-Mass Problems” – back of mol-mol worksheet Lab: Mass Relations in Chemical Reactions (Baking soda & HCl reaction)

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Stoichiometry: Volume-Volume Problems Liquid amounts often measured in volumes. You might use density and molar mass for these problems p. 308 Toolkit Density unit g/mL means ____g = 1 mL Gases 22.41 L = 1 mol of any gas Basics the same: Change to moles, use mole ratio, and then change to desired units.

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Density Solving Volume-Volume Problems Molar mass Mass Known Molar Ratio Mol Known Molar Mass Mol Unknown Volume Known Volume Unknown

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Sample Problem C p. 309 What volume of H 3 PO 4 forms when 56 mL POCl 3 completely react? (density of POCl 3 = 1.67 g/mL; density of H 3 PO 4 = 1.83 g/mL) POCl 3 (l) + 3H 2 O(l) → H 3 PO 4 (l) + 3HCl (g) 56mL POCl 3 x 1.67 g POCl 3 x 1 mol POCl 3 x 1 mol H 3 PO 4 1 mL 153.32g 1 mol POCl 3 X 98.00 g H 3 PO 4 x 1 mL H 3 PO 4 = 33 mL H 3 PO 4 1 mol 1.83 g

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Practice p. 309 # 1-4 C 5 H 12 (l) → C 5 H 8 (l) + 2H 2 (g) Densities: C 5 H 12 = 0.620 g/mL C 5 H 8 = 0.681 g/mL H 2 = 0.0899 g/L = 0.0899 g/1000mL = 8.99 x 10 -5 g/mL Calculate Molar Masses: C 5 H 12 = 72.17 g/mol C 5 H 8 = 68.13 g/mol H 2 = 2.02 g/mol

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1.How many mL of C 5 H 8 can be made from 366mL C 5 H 12 ? 366mL C 5 H 12 x 0.620 g x 1 mol C 5 H 12 x 1 mol C 5 H 8 1 mL 72.17 g 1 mol C 5 H 12 X 68.13 g x 1 mL 1 mol C 5 H 8 0.681 g = 315 mL C 5 H 8

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2. How many liters of H 2 can form when 4.53 x 10 3 mL C 5 H 8 form? 4.53 x 10 3 mLC 5 H 8 x 0.681 g x 1 mol C 5 H 8 x 2 mol H 2 1 mL 68.13 g 1 mol C 5 H 8 X 2.02 g x 1 L 1 mol H 2 0.0899 g = 2030 L H 2

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3. How many mL of C 5 H 12 are needed to make 97.3 mL of C 5 H 8 ? 97.3 mL C 5 H 8 x 0.681 g x 1 mol C 5 H 8 x 1 mol C 5 H 12 1 mL 68.13 g 1 mol C 5 H 8 X 72.17 g x 1 mL 1 mol C 5 H 12 0.620 g = 113 mL C 5 H 12

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4. How many milliliters of H 2 can be made from 1.98 x 10 3 mL C 5 H 12 ? 1.98 x 10 3 mL C 5 H 12 x 0.620 g x 1 mol C 5 H 12 x 2 mol H 2 1 mL 72.17 g 1 mol C 5 H 12 X 2.02 g x 1 mL 1 mol H 2 0.0000899 g = 7.64 x 10 5 mL H 2

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Homework WS “Stoichiometry: Volume-Volume Problems” front only WS “Stoichiometry: MixedProblems” back – classwork 9.1 QUIZ will be on __________________

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9.2 Limiting Reactants and Percentage Yield In this section you will: Identify the limiting reactant for a reaction and use it to calculate theoretical yield. Perform calculations involving percentage yield.

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Limiting Reactant To drive a car you need gasoline and oxygen from the air. When the gas runs out, you can’t go any farther even though there is still plenty of oxygen. The gasoline limits the distance you can travel because it runs out first. Page 312/313 Figures 3&4 Which of the starting supplies limited the number of mums that could be made? Which items where in excess/left over?

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Limiting reactant = substance that controls the quantity of product that can form in a chemical reaction; runs out first Excess reactant = substance that is not used up completely in a reaction. Identify the Limiting Reactant from Lab Data Calculate the amount of product that each could form. Whichever reactant would produce the least amount of product is the limiting reactant.

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Theoretical yield = maximum amount of product that can be made if everything about the reaction works out perfectly; determined by the limiting reactant. Whenever a problem gives you quantities of 2 or more reactants, you must 1.Determine the limiting reactant (makes less product) 2.Use the limiting reactant quantity to determine the theoretical yield.

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Sample E p. 314 Identify the limiting reactant and the theoretical yield of phosphorous acid, H 3 PO 3 if 225g of PCl 3 is mixed with 123 g of H 2 O. PCl 3 + 3H 2 O → H 3 PO 3 + 3HCl Determine how many g of H 3 PO 4 that each reactant could make. Need to calculate molar masses to use as conversion factors.

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225g PCl 3 x 1mol PCl 3 X 1mol H 3 PO 4 x 82.00g H 3 PO 4 137.32g 1 mol PCl 3 1 mol = 134 g H 3 PO 4 123g H 2 O x 1mol PCl 3 X 1mol H 3 PO 4 x 82.00g H 3 PO 4 18.02g 3 mol H 2 O 1 mol = 187 g H 3 PO 4 PCl 3 is the limiting reactant because 134 g is less than 187 g. The theoretical yield is 134 grams of H 3 PO 3.

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Practice p. 314 #1-3 PCl 3 + 3H 2 O → H 3 PO 3 + 3HCl Identify the limiting reactant and the theoretical yield (in grams) of HCl for each pair 1.3.00 mol PCl 3 and 3.00 mol H 2 O 3.00 mol PCl 3 x 3 mol HCl = 9.00 mol HCl 1 mol PCl 3 3.00 mol H 2 Ox 3 mol HCl = 3.00 mol HCl 3 mol H 2 O H 2 O is limiting reactant. 3.00 mol HCl x 36.51 g = 109 grams HCl 1 mol theoretical yield (made)

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Practice p. 314 #1-3 PCl 3 + 3H 2 O → H 3 PO 3 + 3HCl Identify the limiting reactant and the theoretical yield (in grams) of HCl for each pair 2.75.0 g PCl 3 and 75.0 g H 2 O 75.0g PCl 3 x 1 mol PCl 3 x 3 mol HCl = 1.65 mol HCl 136.5 g 1 mol PCl 3 75.0 g H 2 O x 1 mol H 2 O x 3 mol HCl = 4.16 mol HCl 18.023 mol H 2 O PCl 3 is limiting reactant. 1.65 mol HCl x 36.51 g = 60.2 grams HCl 1 mol theoretical yield (made)

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Practice p. 314 #1-3 PCl 3 + 3H 2 O → H 3 PO 3 + 3HCl Identify the limiting reactant and the theoretical yield (in grams) of HCl for each pair 3. 1.00 mol PCl 3 and 50.0 g H 2 O 1.00 mol PCl 3 x 3 mol HCl = 3.00 mol HCl 1 mol PCl 3 50.0 g H 2 O x 1 mol H 2 O x 3 mol HCl = 2.77 mol H 3 PO 3 18.023 mol H 2 O H 2 O is limiting reactant. 2.77 mol HCl x 36.51 g = 101 grams HCl 1 mol theoretical yield (made)

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Limiting Reactants & Industry The most expensive chemicals are chosen as the limiting reactants Less expense reactants can be used in excess to ensure all of the expensive chemicals are completely used up (none wasted).

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Actual Yield and Percentage Yield Although equations tell you what should happen, they cannot always tell you what will happen in real life/in the lab. Some reactions do not make all of the product predicted by the theoretical yield. ACTUAL YIELD = the mass of product actually formed, measured in lab, often less than expected (theoretical).

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Examples of things that could reduce yield: Incomplete distillation/purification needed to separate product from a mixture. Side reactions that can use up reactants without making desired products

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PERCENTAGE YIELD = ratio relating the actual to the theoretical yield; describes how efficient the reaction was. Percentage Yield = actual x 100 theoretical SHOULD ALWAYS BE LESS THAN 100%. If not, you switched the actual & theoretical in the formula!!!

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Sample Problem F p. 317 N 2 + 3 H 2 → 2NH 3 Determine the limiting reactant, theoretical yield and percentage yield if 14.0g N 2 are mixed with 9.0 g H 2 and 16.1 g NH 3 form. 14.0g N 2 x 1 mol N 2 x 2 mol NH 3 x 17.04 g NH 3 = 17.0g NH 3 28.02 g 1 mol N 2 1 mol NH 3 9.0g H 2 x 1 mol N 2 x 2 mol NH 3 x 17.04 g NH 3 = 51 g NH 3 2.02 g 3 mol H 2 1 mol NH 3 N 2 makes the smaller amount and is the limiting reactant.

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Sample Problem F p. 317 N 2 + 3 H 2 → 2NH 3 Determine the limiting reactant, theoretical yield and percentage yield if 14.0g N 2 are mixed with 9.0 g H 2 and 16.1 g NH 3 form. N 2 makes the smaller amount and is the limiting reactant. The theoretical amount made is 17.0g NH 3. The actual amount made is 16.1 g NH 3 PERCENTAGE YIELD = 16.1 g x 100 = 94.7% 17.0g

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Practice p. 317 #1 N 2 + 3 H 2 → 2NH 3 1. Determine the limiting reactant, theoretical yield and percentage yield if 14.0g N 2, 3.15g H 2 and actual is 14.5 g NH 3. 14.0g N 2 x 1 mol N 2 x 2 mol NH 3 x 17.04 g NH 3 = 17.0g NH 3 28.02 g 1 mol N 2 1 mol NH 3 3.15g H 2 x 1 mol N 2 x 2 mol NH 3 x 17.04 g NH 3 = 17.7 g NH 3 2.02 g 3 mol H 2 1 mol NH 3 N 2 makes the smaller amount and is the limiting reactant.

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N 2 + 3 H 2 → 2NH 3 Determine the limiting reactant, theoretical yield and percentage yield if 14.0g N 2 are mixed with 9.0 g H 2 and 16.1 g NH 3 form. N 2 makes the smaller amount and is the limiting reactant. The theoretical amount made is 17.0g NH 3. The actual amount made is 14.5 g NH 3 PERCENTAGE YIELD = 14.5 g x 100 = 85.3% 17.0g

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Homework Section 9.2 Review p. 319 # 1,3,4,5,6,8,10,12,14 Quiz on 9.2 will be on __________________.

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