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Chapter 11: Stoichiometry

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Chapter 11: Stoichiometry Stoichiometry BIG Idea Mass relationships in chemical reactions confirm the law of conservation of mass. 11.1 Defining Stoichiometry Idea MAIN The amount of each reactant present at the start of a Carbon dioxide and water chemical reaction determines how much product can form. 11.2 Stoichiometric Calculations MAIN Idea The solution to every stoichiometric problem requires a balanced chemical equation. 11.3 Limiting Reactants MAIN Idea A chemical reaction stops when one of the reactants is used up. 11.4 Percent Yield MAIN Idea Percent yield is a measure of the efficiency of a chemical reaction. ChemFacts • Green plants make their own food Chloroplast through photosynthesis. • Photosynthesis occurs within structures called chloroplasts in the cells of plants. • The balanced chemical equation for the photosynthesis is: 6C O2 + 6 H 2 O → C 6 H 12 O 6 + 6 O 2 • On a summer day, one acre of corn produces enough oxygen (a product of photosynthesis) to meet the respiratory needs of 130 people. 366 ©CLIVE SCHAUPMEYER/AGSTOCKUSA/SCIENCE PHOTO LIBRARY/Photo Researchers Inc. SStart-Uptart-Up AActivitiesctivities Steps in Stoichiometric LAUNNCHCH LLabab Calculations Make the following What evidence can you observe Foldable to help you summarize the that a reaction is taking place? steps in solving a stoichiometric problem. During a chemical reaction, reactants are consumed as new products are formed. Often, there are several telltale STEP 1 Fold a sheet signs that a chemical reaction is taking place. of paper in half lengthwise. Procedure 1. Read and complete the lab safety form. 2. Use a 10-mL graduated cylinder to measure out 5.0 mL 0.01M potassium permanganate (KMn O4 ). Add the solution to a 100-mL beaker. STEP 2 Fold in half 3. Clean and dry the graduated cylinder, and then use widthwise and then in half again. it to measure 5.0 mL 0.01M sodium hydrogen sulfite solution (NaHS O 3 ). Slowly add this solution to the beaker while stirring with a stirring rod. Record your observations. STEP 3 Unfold and 4. cut along the folds of Repeat Step 3 until the KMn O4 solution in the beaker turns colorless. Stop adding the NaHS O3 solution as the top flap to make soon as you obtain a colorless solution. Record your four tabs. 1. observations. 2. Analysis STEP 4 Label the tabs 3. 1. Identify the evidence you observed that a chemical with the steps in stoichiometric reaction was occurring. calculations. 4. 2. Explain why slowly adding the NaHS O 3 solution while stirring is a better experimental technique than adding &/,$!",%3 Use this Foldable with Section 11.2. As 5.0 mL of the solution all at once. you read this section, summarize each step on a tab and Inquiry Would anything more have happened if you include an example of the step. continued to add NaHS O3 solution to the beaker? Explain. Visit glencoe.com to: ▶ study the entire chapter online ▶ explore ▶ take Self-Check Quizzes ▶ use the Personal Tutor to work Example Problems step-by-step ▶ access Web Links for more information, projects, and activities ▶ find the Try at Home Lab, Baking Soda Stoichiometry Chapter 11 • Stoichiometry 367 Section 111.11.1 Objectives Defining Stoichiometry ◗ Describe the types of relationships indicated by a balanced chemical MAIN Idea The amount of each reactant present at the start of a equation. chemical reaction determines how much product can form. ◗ State the mole ratios from a balanced chemical equation. Real-World Reading Link Have you ever watched a candle burning? You might have watched the candle burn out as the last of the wax was used up. Review Vocabulary Or, maybe you used a candle snuffer to put out the flame. Either way, when the reactant: the starting substance in a candle stopped burning, the combustion reaction ended. chemical reaction Particle and Mole Relationships New Vocabulary In doing the Launch Lab, were you surprised when the purple color of stoichiometry potassium permanganate disappeared as you added sodium hydrogen mole ratio sulfite? If you concluded that the potassium permanganate had been used up and the reaction had stopped, you are right. Chemical reactions stop when one of the reactants is used up. When planning the reaction of potassium permanganate and sodium hydrogen sulfite, a chemist might ask, “How many grams of potassium permanganate are needed to react completely with a known mass of sodium hydrogen sulfite?” Or, when analyzing a photosynthesis reaction, you might ask, “How much oxygen and carbon dioxide are needed to form a known mass of sugar.” Stoichiometry is the tool for answering these questions. Stoichiometry The study of quantitative relationships between the amounts of reactants used and amounts of products formed by a chemi- cal reaction is called stoichiometry. Stoichiometry is based on the law of conservation of mass. Recall from Chapter 3 that the law states that matter is neither created nor destroyed in a chemical reaction. In any chemical reaction, the amount of matter present at the end of the reac- tion is the same as the amount of matter present at the beginning. Therefore, the mass of the reactants equals the mass of the products. Note the reaction of powdered iron (Fe) with oxygen ( O2 ) shown in Figure 11.1. Although iron reacts with oxygen to form a new com- pound, iron(III) oxide (F e2 O 3 ), the total mass is unchanged. ■ Figure 11.1 The balanced chemi- cal equation for this reaction between iron and oxygen provides the relation- ships between amounts of reactants and products. 368 Chapter 11 • Stoichiometry ©Charles D. Winters/Photo Researchers, Inc. Interactive Table Relationships Explore Table balanced chemical equations Derived from a at glencoe.com. 11.1 Balanced Chemical Equation 4Fe(s) + 3O 2 (g) → 2Fe 2 O 3 (s) iron + oxygen → iron(III) oxide 4 atoms Fe + 3 molecules O2 → 2 formula units F e2 O 3 4 mol Fe + 3 mol O2 → 2 mol F e 2 O 3 223.4 g Fe + 96.00 g O2 → 319.4 g F e 2 O 3 319.4 g reactants → 319.4 g products The balanced chemical equation for the chemical reaction shown in Figure 11.1 is as follows. 4Fe(s) + 3 O 2 (g) → 2F e 2 O 3 ( s ) You can interpret this equation in terms of representative particles by saying that four atoms of iron react with three molecules of oxygen to produce two formula units of iron(III) oxide. Remember that coeffi- cients in an equation represent not only numbers of individual particles but also numbers of moles of particles. Therefore, you can also say that four moles of iron react with three moles of oxygen to produce two moles of iron(III) oxide. The chemical equation does not directly tell you anything about the VOCABULARY masses of the reactants and products. However, by converting the WORD ORIGIN known mole quantities to mass, the mass relationships become obvious. Stoichiometry Recall that moles are converted to mass by multiplying by the molar comes from the Greek words mass. The masses of the reactants are as follows. stoikheion, which means element, and metron, which means to 55.85 g Fe 4 mol Fe × _ = 223.4 g Fe measure 1 mol Fe 32.00 g O _2 3 mol O 2 × = 96.00 g O 2 1 mol O2 The total mass of the reactants is: (223.4 g + 96.00 g) = 319.4 g Similarly, the mass of the product is calculated as follows: 159.7 g F e O __2 3 2 mol F e 2 O 3 × = 319.4 g 1 mol F e2 O 3 Note that the mass of the reactants equals the mass of the product. mass of reactants = mass of products 319.4 g = 319.4 g As predicted by the law of conservation of mass, the total mass of the reactants equals the mass of the product. The relationships that can be determined from a balanced chemical equation are summarized in Table 11.1. Reading Check List the types of relationships that can be derived from the coefficients in a balanced chemical equation. Section 11.1 • Defining Stoichiometry 369 EXAMPLE Problem 11.1 Math Handbook Interpreting Chemical Equations The combustion of propane ( C 3 H 8 ) provides Rounding energy for heating homes, cooking food, and soldering metal parts. Interpret the page 952 equation for the combustion of propane in terms of representative particles, moles, and mass. Show that the law of conservation of mass is observed. 1 Analyze the Problem The coefficients in the balanced chemical equation shown below represent both moles and representative particles, in this case molecules. Therefore, the equation can be interpreted in terms of molecules and moles. The law of conservation of mass will be verified if the masses of the reactants and products are equal. Known C 3 H 8 (g) + 5 O 2 (g) → 3C O 2 (g) + 4 H 2 O(g) Unknown Equation interpreted in terms of molecules = ? Equation interpreted in terms of moles = ? Equation interpreted in terms of mass = ? 2 Solve for the Unknown The coefficients in the chemical equation indicate the number of molecules. 1 molecule C3H8 + 5 molecules O2 → 3 molecules CO2 + 4 molecules H2O The coefficients in the chemical equation also indicate the number of moles. 1 mol C3H8 + 5 mol O2 → 3 mol CO2 + 4 mol H2O To verify that mass is conserved, first convert moles of reactant and product to mass by multiplying by a conversion factor—the molar mass—that relates grams to moles. grams reactant or product moles of reactant or product × ___ = grams of reactant or product 1 mol reactant or product 44.09 g C H __3 8 1 mol C 3 H 8 × = 44.09 g C 3 H 8 Calculate the mass of the reactant C3 H8 .
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