Limiting Reactant - Bellwork: Why doesn’t the warm air...

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Unformatted text preview: Bellwork: Why doesn’t the warm air from a lit birthday candle rise within an orbiting space station? Why does the candle burn for only a few moments before it extinguishes itself? Warm air rises because it is less dense than the surrounding air. Within an orbiting space station, however, there is no up or down, hence, no where for the warm air to rise. It instead remains hovered around the flame. But this warm air contains the products of combustion—primarily water vapor and carbon dioxide—which eventually smother the flame. OBJECTIVES 3.3.8 Describe a method for determining which of two reactants is a limiting reactant. 3.3.9 Calculate the amount in moles or mass in grams of a product, given the amounts in moles or masses in grams of two reactants, one of which is in excess. Stoichiometry – Ch. 9 II. Stoichiometry in the Real World Real (p. 288-294) Limiting Reactant Limiting Reactant Limiting Reactants Limiting • Limiting Reactant – used up in a reaction – determines the amount of product • Excess Reactant – added to ensure that the other reactant is completely used up – cheaper & easier to recycle Consider the following Analogy: 2 Wheels + 1 Body + 1 Handle bar + 1 Gear Chain = 1 bike How many bikes can be produced given, 8 wheels, 5 bodies, 6 handle bar and 5 gear chain? Bike Analogy Bike Analogy Consider the following Analogy: 2 Wheels + 1 Body + 1 Handle bar + 1 Gear Chain = 1 bike How many bikes can be produced given, 8 wheels, 5 bodies, 6 handle bar and 5 gear chain? Limiting Reactant Excess Reactant Bike Analogy Bike Analogy Consider the following Analogy: Cheeseburger Analogy Cheeseburger Analogy 2 Cheese + 1 burger patty + 1 bun = cheese burger Given three hamburger patties, six buns, and 12 slices of cheese, how many cheese burger can be made? Consider the following Analogy: Cheeseburger Analogy Cheeseburger Analogy LR 2 Cheese + 1 burger patty + 1 bun = 1 cheese burger Given three hamburger patties, six buns, and 12 slices of cheese, how many cheese burger can be made? ER Limiting Reactants Limiting 1. Write a balanced equation. 2. For each reactant, calculate the product formed. 3. Smaller answer indicates: limiting reactant amount of product amount of Silicon dioxide (quartz) is usually quite unreactive but reacts readily Silicon dioxide (quartz) is usually quite unreactive but reacts readily with hydrogen fluoride according to the following equation. LR SiO2(s) + 4HF(g) 4.5 mol 2.0 mol Sample Problem 9­6 SiF4(g) + 2H2O(l) ? mol If 2.0 mol of HF are exposed to 4.5 mol of SiO2, which is the limiting reactant? LR 1 mol SiO2 2.0 mol HF x 4 mol HF = (smaller number) 4.5 mol SiO2 x 4 mol HF 1 mol SiO2 = 18 mol HF Methanol, CH3OH, is the simplest of the alcohols. It is synthesized by Methanol, CH the reaction of hydrogen and carbon monoxide. CO(g) + 2H2(g) 500 mol 750 mol Additional Sample Problem #1 CH3OH(g) ? mol a. If 500 mol CO react with 750 mol of H2, which is the limiting reactant? b. How many moles of CH3OH are formed? c. How many moles of excess reactant remain unchanged? Methanol, CH3OH, is the simplest of the alcohols. It is synthesized by Methanol, CH the reaction of hydrogen and carbon monoxide. LR CO(g) + 2H2(g) 500 mol 750 mol Additional Sample Problem #1 CH3OH(g) ? mol a. If 500 mol CO react with 750 mol of H2, which is the limiting reactant? 2 mol H2 500 mol CO x 1 mol CO LR 750 mol H2 x 1 mol CO 2 mol H2 = = 375 mol CO Methanol, CH3OH, is the simplest of the alcohols. It is synthesized by Methanol, CH the reaction of hydrogen and carbon monoxide. LR CO(g) + 2H2(g) 500 mol 750 mol Additional Sample Problem #1 CH3OH(g) ? mol b. How many moles of CH3OH are formed? LR 750 mol H 2 1 mol CH3OH x 2 mol H2 = Methanol, CH3OH, is the simplest of the alcohols. It is synthesized by Methanol, CH the reaction of hydrogen and carbon monoxide. LR CO(g) + 2H2(g) 500 mol 750 mol Additional Sample Problem #1 CH3OH(g) ? mol b. How many moles of excess reactant remain unchanged? 750 mol H2 LR 750 mol H 2 x 1 mol CO 2 mol H2 = 375 mol CO 500 mol CO originally present – 375 mol CO consumed = 125 mol remaining 1 mol CH3OH x 2 mol H2 = The black oxide of iron, Fe3O4, occurs in nature as the mineral The black oxide of iron, Fe magnetite. This substance can also be made in the laboratory by the reaction between red­hot iron and steam according to the following equation. 3Fe(s) + 4H2O(g) 167 g 36.0 g Sample Problem 9­7 Fe3O4(g) + 4H2(g) ? g a. If 36.0 g of H2O is mixed with 167 g of Fe, which is the limiting reactant? b. What mass in grams of black iron oxide is produced? c. What mass in grams of excess reactant remains when the reaction is complete? The black oxide of iron, Fe3O4, occurs in nature as the mineral The black oxide of iron, Fe magnetite. This substance can also be made in the laboratory by the reaction between red­hot iron and steam according to the following equation. LR Sample Problem 9­7 3Fe(s) + 4H2O(g) 167 g 36.0 g Fe3O4(g) + 4H2(g) ? g a. If 36.0 g of H2O is mixed with 167 g of Fe, which is the limiting reactant? LR 1 mol H2O 3 mol Fe 2 36.0 g H O x x 18.02 g H2O 4 mol H2O 1 mol Fe 55.85 g Fe 4 mol H2O x 3 mol Fe x 55.85 g Fe 1 mol Fe = 83.78 g Fe 167 g Fe x 18.02 g H2O x 1 mol H2O = The black oxide of iron, Fe3O4, occurs in nature as the mineral The black oxide of iron, Fe magnetite. This substance can also be made in the laboratory by the reaction between red­hot iron and steam according to the following equation. LR Sample Problem 9­7 3Fe(s) + 4H2O(g) 167 g 36.0 g Fe3O4(g) + 4H2(g) ? g b. What mass in grams of black iron oxide is produced? LR 1 mol H2O 1 mol Fe3O4 231.55 g Fe3O4 2 36.0 g H O x x x = 2 2 34 18.02 g H O 4 mol H O 1 mol Fe O Limiting Reactants Limiting 3Fe(s) + 4H2O(g) XS LR Fe3O4(g) + 4H2(g) Limiting reactant: H2O Excess reactant: Fe Product Formed: 116 g Fe3O4 left over iron The black oxide of iron, Fe3O4, occurs in nature as the mineral The black oxide of iron, Fe magnetite. This substance can also be made in the laboratory by the reaction between red­hot iron and steam according to the following equation. LR Sample Problem 9­7 3Fe(s) + 4H2O(g) 167 g 36.0 g Fe3O4(g) + 4H2(g) ? g c. What mass in grams of excess reactant remains when the reaction is complete? LR 1 mol H2O 1 mol Fe3O4 231.55 g Fe3O4 2 36.0 g H O x x x = 2 2 34 18.02 g H O 4 mol H O 1 mol Fe O 1 mol H2O 3 mol Fe x x 18.02 g H2O 4 mol H2O 55.85 g Fe 1 mol Fe 36.0 g H O 2 x = 83.8 g Fe 83.2 g Fe remaining 167 g Fe originally present ­ 83.8 g Fe consumed = Homework Homework Section 9-2 pg. 289 Practice Problems #1-2 pg. 291 Practice Problems#1-2 Due: T 1/15 ...
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This note was uploaded on 11/30/2010 for the course CHEM 121 taught by Professor Heske during the Spring '08 term at University of Nevada, Las Vegas.

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