Chapter 12 Stoichiometry 12.1 The Arithmetic of Equations

Chapter 12 Stoichiometry 12.1 The Arithmetic of Equations
12.2 Chemical Calculations 12.3 Limiting Reagent and Percent Yield Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CHEMISTRY & YOU How do you figure out how much starting material you need to make a finished product? When making bikes, you need parts like wheels, handlebars, pedals, and frames. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

How do chemists use balanced chemical equations?
Using Equations Using Equations How do chemists use balanced chemical equations? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Using Equations Everyday Equations Travel Time Tricycle company produces 640 tricycles each week. How can you determine the number of parts they need per week? + Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Using Equations Everyday Equations Travel Time Tricycle company produces 640 tricycles each week. How can you determine the number of parts they need per week? The major components are the frame (F), the seat (S), the wheels (W), the handlebars (H), and the pedals (P). + Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Using Equations Everyday Equations Travel Time Tricycle company produces 640 tricycles each week. How can you determine the number of parts they need per week? The finished tricycle, your product, has a “formula” of FSW3HP2. + Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Using Equations Everyday Equations Travel Time Tricycle company produces 640 tricycles each week. How can you determine the number of parts they need per week? The balanced equation for making a single tricycle is: F + S + 3W + H + 2P  FSW3HP2 + Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Using a Balanced Equation as a Recipe
Sample Problem 12.1 Using a Balanced Equation as a Recipe In a five-day workweek, Travel Time is scheduled to make 640 tricycles. How many wheels should be in the plant on Monday morning to make these tricycles? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Analyze List the knowns and the unknown.
Sample Problem 12.1 Analyze List the knowns and the unknown. 1 Use the balanced equation to identify a conversion factor that will allow you to calculate the unknown. The conversion you need to make is from tricycles (FSW3HP2) to wheels (W). UNKNOWN number of wheels = ? wheels KNOWNS number of tricycles = 640 tricycles = 640 FSW3HP2 F + S + 3W + H + 2P  FSW3HP2 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Calculate Solve for the unknown.
Sample Problem 12.1 Calculate Solve for the unknown. 2 Identify a conversion factor that relates wheels to tricycles. You can write the two conversion factors relating wheels to tricycles. 3 W 1FSW3HP2 and 1 FSW3HP2 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.1 Calculate Solve for the unknown. 2 The desired unit is W; so use the conversion factor on the left. Multiply the number of tricycles by the conversion factor. 3 W 1 FSW3HP2 and 640 FSW3HP2  = 1920 W 3 W 1 FSW3HP2 When using conversion factors, remember to cross out like units when they are in both the numerator and denominator. This will help you check that you are using the correct conversion factor. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Evaluate Does the result make sense?
Sample Problem 12.1 Evaluate Does the result make sense? 3 If three wheels are required for each tricycle and more than 600 tricycles are being made, then a number of wheels in excess of 1800 is a logical answer. The unit of the known (FSW3HP2) cancels. The answer has the correct unit (W). Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Balanced Chemical Equations
Using Equations Balanced Chemical Equations Chemists use balanced chemical equations as a basis to calculate how much reactant is needed or how much product will be formed in a reaction. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Using Equations Balanced Chemical Equations Chemists use balanced chemical equations as a basis to calculate how much reactant is needed or how much product will be formed in a reaction. When you know the quantity of one substance in a reaction, you can calculate the quantity of any other substance consumed or created in the reaction. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Using Equations Balanced Chemical Equations The calculations of quantities in chemical reactions is a subject of chemistry called stoichiometry. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Using Equations Balanced Chemical Equations The calculations of quantities in chemical reactions is a subject of chemistry called stoichiometry. For chemists, stoichiometry is a form of bookkeeping. It allows chemists to tally the amounts of reactants and products using ratios of moles or representative particles. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Cayla is using a recipe to make chocolate chip cookies
Cayla is using a recipe to make chocolate chip cookies. She wants to double the number of cookies that the recipe will make. The original recipe calls for 2 cups of chocolate chips. How many cups of chips should Cayla use for a double recipe? A. 2 cups C. 1 cup B. 4 cups D. 8 cups Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Chemical Equations Ammonia is produced industrially by the reaction of nitrogen with hydrogen. N2(g) + 3H2(g)  2NH3(g) The balanced chemical equation tells you the relative amounts of reactants and product in the reaction. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Chemical Equations Ammonia is produced industrially by the reaction of nitrogen with hydrogen. N2(g) + 3H2(g)  2NH3(g) The balanced chemical equation tells you the relative amounts of reactants and product in the reaction. Your interpretation of the equation depends on how you quantify the reactants and products. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CHEMISTRY & YOU How can you determine the amount of each reactant you need to make a product? A recipe or an equation, such as a balanced chemical equation, is used to determine how much starting material is needed or how much product will be made. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Chemical Equations A balanced chemical equation can be interpreted in terms of different quantities, including numbers of atoms, molecules, or moles; mass; and volume. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

2 atoms N + 6 atoms H 2 atoms N and 6 atoms H
Chemical Equations Number of Atoms At the atomic level, a balanced equation indicates the number and types of atoms that are rearranged to make the product or products. In the synthesis of ammonia, the reactants are composed of two atoms of nitrogen and six atoms of hydrogen. 2 atoms N + 6 atoms H atoms N and 6 atoms H Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Chemical Equations Number of Molecules Nitrogen and hydrogen will always react to form ammonia in a 1:3:2 ratio of molecules. It is not practical to count very small numbers of molecules and allow them to react. You could take Avogadro’s number (6.02  1023 molecules) of nitrogen molecules and make them react with three times Avogadro’s number of hydrogen molecules. 1  ( ) + 3  ( )  2  ( ) 6.02  1023 molecules N2 6.02  1023 molecules H2 6.02  1023 molecules NH3 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Chemical Equations Moles Since a balanced chemical equation tells you the number of representative particles, it also tells you the number of moles. In the synthesis of ammonia, one mole of nitrogen molecules reacts with three moles of hydrogen molecules to form two moles of ammonia molecules. 1 mol N2 + 3 mol H2  2 mol NH3 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Chemical Equations Mass A balanced chemical equation obeys the law of conservation of mass. Mass can be neither created nor destroyed in an ordinary chemical or physical process. The total mass of the atoms in a reaction does not change. 28 g N2 + (3  2 g H2)  (2  17 g NH3) 28 g N2 + 6 g H2  34 g NH3 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Chemical Equations Volume If you assume standard temperature and pressure, the balanced chemical equation also tells you about the volumes of gases. 1 mol of any gas at STP occupies a volume of 22.4 L. 22.4 L N2 + (3  22.4 L H2)  (2  22.4 L NH3) 22.4 L N L H2  44.8 L NH3 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

2H2S(g) + 3O2(g)  2SO2(g) + 2H2O(g)
Sample Problem 12.2 Interpreting a Balanced Chemical Equation Hydrogen sulfide, which smells like rotten eggs, is found in volcanic gases. The balanced equation for the burning of hydrogen sulfide is: 2H2S(g) + 3O2(g)  2SO2(g) + 2H2O(g) Interpret this equation in terms of a. numbers of representative particles and moles. b. masses of reactants and produces. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Analyze Identify the relevant concepts.
Sample Problem 12.2 Analyze Identify the relevant concepts. 1 The coefficients in the balanced equation give the relative number of representative particles and moles of reactants and products. A balanced chemical equation obeys the law of conservation of mass. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Solve Apply concepts to this situation.
Sample Problem 12.2 Solve Apply concepts to this situation. 2 Use the coefficients in the balanced equation to identify the number of representative particles and moles. 2 molecules H2S + 3 molecules O2  2 molecules SO2 + 2 molecules H2O 2 mol H2S + 3 mol O2  2 mol SO2 + 2 mol H2O Remember that atoms and molecules are both representative particles. In this equation, all the reactants and products are molecules; so all the representative particles are molecules. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Solve Apply concepts to this situation.
Sample Problem 12.2 Solve Apply concepts to this situation. 2 Use the periodic table to calculate the molar mass of each reactant and product. 1 mol H2S = 34.1 g H2S 1 mol O2 = 32.0 g O2 1 mol SO2 = 64.1 g/mol SO2 1 mol H2O = 18.0 g/mol H2O Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

2 mol H2S + 3 mol O2  2 mol SO2 + 2 mol H2O
Sample Problem 12.2 Solve Apply concepts to this situation. 2 Multiply the number of moles of each reactant and product by its molar mass. 2 mol H2S + 3 mol O2  2 mol SO2 + 2 mol H2O (2 mol  ) + (3 mol  )  g mol (2 mol  ) + (2 mol  ) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

2 mol H2S + 3 mol O2 2 mol SO2 + 2 mol H2O
Sample Problem 12.2 Solve Apply concepts to this situation. 2 Multiply the number of moles of each reactant and product by its molar mass. 2 mol H2S + 3 mol O mol SO2 + 2 mol H2O (2 mol  ) + (3 mol  )  g mol (2 mol  ) + (2 mol  ) g mol 68.2 g H2S O2  g SO g H2O 164.2 g = g Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Chemical Equations The table below summarizes the information derived from the balanced chemical equation for the formation of ammonia. N2(g) + 2H2(g) 2NH3(g) 2 atoms N 6 atoms H 2 atoms N and 6 atoms H 1 molecule N2 3 molecules H2 2 molecules NH3 10 molecules N2 30 molecules H2 20 molecules NH3 1  3  2  1 mol N2 3 mol H2 2 mol NH3 23 g N2 3  2 g H2 2  17 g NH3 34 g reactants 34 g products Assume STP 22.4 L N2 67.2 L H2 44.8 L NH3 6.02  1023 molecules N2 ( ) molecules H2 molecules NH2 22.4 L Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Mass and atoms are conserved in every chemical reaction.
Chemical Equations Mass and atoms are conserved in every chemical reaction. Molecules, formula units, moles, and volumes are not necessarily conserved—although they may be. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

2H2(g) + 2NO(g)  N2(g) + 2H2O(g)
Interpret the following equation in terms of volumes of gas at STP. 2H2(g) + 2NO(g)  N2(g) + 2H2O(g) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Interpret the following equation in terms of volumes of gas at STP.
2H2(g) + 2NO(g)  N2(g) + 2H2O(g) 44.8 L H2(g) L NO(g)  22.4 L N2(g) L H2O(g) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Mass and atoms are conserved in every chemical reaction.
Key Concepts A balanced chemical equation provides the same kind of quantitative information that a recipe does. Chemists use balanced chemical equations as a basis to calculate how much reactant is needed or product is formed in a reaction. A balanced chemical equation can be interpreted in terms of different quantities, including numbers of atoms, molecules, or moles; mass; and volume. Mass and atoms are conserved in every chemical reaction. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Glossary Terms stoichiometry: that portion of chemistry dealing with numerical relationships in chemical reactions; the calculation of quantities of substances involved in chemical equations Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

The Mole and Quantifying Matter
BIG IDEA The Mole and Quantifying Matter Balanced chemical equations are the basis for stoichiometric calculations. The coefficients of a balanced equation indicate the number of particles, mole, or volumes of gas in the reaction. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Stoichiometry Chapter 12 12.2 Chemical Calculations
12.1 The Arithmetic of Equations 12.2 Chemical Calculations 12.3 Limiting Reagent and Percent Yield Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

How do manufacturers know how to make enough of their desired product?
CHEMISTRY & YOU How do manufacturers know how to make enough of their desired product? If chemical plants produce too much ammonia, then it might be wasted. But if too little is produced, then there might not be enough for all their customers. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Writing and Using Mole Ratios
How are mole ratios used in chemical calculations? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

A mole ratio is a conversion factor derived from the coefficients of a balanced chemical equation interpreted in terms of moles. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

In chemical calculations, mole ratios are used to convert between a given number of moles of a reactant or product to moles of a different reactant or product. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Look again at the balanced equation for the production of ammonia. N2(g) + 3H2(g)  2NH3(g) The three mole ratios derived from the balanced equation above are: 2 mol NH3 1 mol N2 3 mol H2 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Mole-Mole Calculations In the mole ratio below, W is the unknown, wanted quantity and G is the given quantity. The values of a and b are the coefficients from the balanced equation. The general solution for a mole-mole problem is given by x mol G  = mol W b mol W xb a mol G a Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Calculating Moles of a Product
Sample Problem 12.3 Calculating Moles of a Product How many moles of NH3 are produced when 0.60 mol of nitrogen reacts with hydrogen? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

moles of nitrogen = 0.60 mol N2 UNKNOWN moles of ammonia = ? mol NH3
Sample Problem 12.3 Analyze List the known and the unknown. 1 N2 + 3H2  2NH3 The conversion is mol N2  mol NH3. 1 mol N2 combines with 3 mol H2 to produce 2 mol NH3. To determine the number of moles of NH3, the given quantity of N2 is multiplied by the form of the mole ratio from the balanced equation that allows the given unit to cancel. KNOWN moles of nitrogen = 0.60 mol N2 UNKNOWN moles of ammonia = ? mol NH3 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.3 Calculate Solve for the unknown. 2 Write the mole ratio that will allow you to convert from moles N2 to moles NH3. 1 mol N2 2 mol NH3 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.3 Calculate Solve for the unknown. 2 Multiply the given quantity of N2 by the mole ratio in order to find the moles of NH3. 0.60 mol N2  = 1.2 mol NH3 2 mol NH3 1 mol N2 Remember that the mole ratio must have N2 on the bottom so that the mol N2 in the mole ratio will cancel with mol N2 in the known. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.3 Evaluate Does the result make sense? 3 The ratio of 1.2 mol NH3 to 0.60 mol N2 is 2:1, as predicted by the balanced equation. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Mass-Mass Calculations In the laboratory, the amount of a substance is usually determined by measuring its mass in grams. If a given sample is measured in grams, then the mass can be converted to moles by using the molar mass. Then the mole ratio from the balanced equation can be used to calculate the number of moles of the unknown. If it is the mass of the unknown that needs to be determined, the number of moles of the unknown can be multiplied by the molar mass. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Steps for Solving a Mass-Mass Problem 1. Change the mass of G to moles of G (mass G  mol G) by using the molar mass of G. mass G  = mol G 1 mol G molar mass G Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Steps for Solving a Mass-Mass Problem 1. Change the mass of G to moles of G (mass G  mol G) by using the molar mass of G. mass G  = mol G 1 mol G molar mass G 2. Change the moles of G to moles of W (mol G  mol W ) by using the mole ratio from the balanced equation. mol G  = mol W b mol W a mol G Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Steps for Solving a Mass-Mass Problem 1. Change the mass of G to moles of G (mass G  mol G) by using the molar mass of G. mass G  = mol G 1 mol G molar mass G 2. Change the moles of G to moles of W (mol G  mol W ) by using the mole ratio from the balanced equation. mol G  = mol W b mol W a mol G 3. Change the moles of W to grams of W (mol W  mass W ) by using the molar mass of W. mol W  = mass W 1 mol W molar mass W Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

The figure below shows another way to represent the steps for doing mole-mass and mass-mole stoichiometric calculations. For a mole-mass problem, the first conversion is skipped. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

The figure below shows another way to represent the steps for doing mole-mass and mass-mole stoichiometric calculations. For a mole-mass problem, the first conversion is skipped. For a mass-mole problem, the last conversion is skipped. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Calculating the Mass of a Product
Sample Problem 12.4 Calculating the Mass of a Product Calculate the number of grams of NH3 produced by the reaction of 5.40 g of hydrogen with an excess of nitrogen. The balanced equation is: N2(g) + 3H2(g)  2NH3(g) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

3 mol H2 = 2 mol NH3(from balanced equation)
Sample Problem 12.4 Analyze List the knowns and the unknown. 1 N2(g) + 3H2(g)  2NH3(g) KNOWNS mass of hydrogen = 5.40 g H2 3 mol H2 = 2 mol NH3(from balanced equation) 1 mol H2 = 2.0 g H2 (molar mass) 1 mol NH3 = 17.0 g NH3 (molar mass) UNKNOWN mass of ammonia = ? g NH3 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.4 Analyze List the knowns and the unknown. 1 The mass of hydrogen will be used to find the mass of ammonia: g H2  g NH3. The coefficients of the balanced equation show that 3 mol H2 reacts with 1 mol N2 to produce 2 mol NH3. The following steps are necessary to determine the mass of the ammonia: g H2  mol H2  mol NH3  g NH3 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Change given unit to moles
Sample Problem 12.4 Calculate Solve for the unknown. 2 Start with the given quantity, and convert from mass to moles. 5.40 g H2  1 mol H2 2.0 g H2 Don’t forget to cancel the units at each step. Given quantity Change given unit to moles Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.4 Calculate Solve for the unknown. 2 Then convert from moles of reactant to moles of product by using the correct mole ratio.  2 mol NH3 3 mol H2 5.40 g H2  1 mol H2 2.0 g H2 Given quantity Change given unit to moles Mole ratio Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.4 Calculate Solve for the unknown. 2 Finish by converting from moles to grams. Use the molar mass of NH3.  2 mol NH3 3 mol H2 5.40 g H2  1 mol H2 2.0 g H2 17.0 g NH3 1 mol NH3 Given quantity Change given unit to moles Mole ratio Change moles to grams = 30.6 g NH3 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.4 Evaluate Does the result make sense? 3 Because there are three conversion factors involved in this solution, it is more difficult to estimate an answer. Because the molar mass of NH3 is substantially greater than the molar mass of H2, the answer should have a larger mass than the given mass. The answer should have two significant figures. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

What mass of phosphorus will be needed to produce 3.25 mol of P4O10?
Phosphorus burns in air to produce a phosphorus oxide in the following reaction: 4P(s) + 5O2(g)  P4O10(s) What mass of phosphorus will be needed to produce 3.25 mol of P4O10? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

= 403 g P 4 mol P 31.0 g P 3.25 mol P4O10  1 mol P4O10 1 mol P
Phosphorus burns in air to produce a phosphorus oxide in the following reaction: 4P(s) + 5O2(g)  P4O10(s) What mass of phosphorus will be needed to produce 3.25 mol of P4O10? 3.25 mol P4O10  = 403 g P 31.0 g P 1 mol P 1 mol P4O10 4 mol P Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Other Stoichiometric Calculations
What is the general procedure for solving a stoichiometric problem? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

In a typical stoichiometric problem: The given quantity is first converted to moles. Then, the mole ratio from the balanced equation is used to calculate the number of moles of the wanted substance. Finally, the moles are converted to any other unit of measurement related to the unit mole, as the problem requires. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Recall that the mole can be related to other quantities: 1 mol = 6.02 x 1023 particles 1 mol of a gas = 22.4 L at STP These provide four more conversion factors: 1 mol 6.02x1023 particles and 1 mol 22.4 L and Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CHEMISTRY & YOU How do you think air bag manufacturers know how to get the right amount of air in an inflated air bag? They take the volume of air needed to inflate the bag and convert it to number of moles (assuming STP). Then, they use the mole ratio from a balanced chemical equation to calculate the number of moles of reactants needed. This could be converted to any other unit of measurement related to the unit mole. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Calculating the Molecules of a Product
Sample Problem 12.5 Calculating the Molecules of a Product How many molecules of oxygen are produced when 29.2 g of water is decomposed by electrolysis according to this balanced equation? 2H2O(l) H2(g) + O2(g) electricity Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

g H2O  mol H2O  mol O2  molecules O2
Sample Problem 12.5 Analyze List the knowns and the unknown. 1 The following calculations need to be performed: g H2O  mol H2O  mol O2  molecules O2 The appropriate mole ratio relating mol O2 to mol H2O from the balanced equation is: 1 mol O2 2 mol H2O Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.5 Analyze List the knowns and the unknown. 1 KNOWNS mass of water = 29.2 g H2O 2 mol H2O = 1 mol O2 (from balanced equation) 1 mol H2O = 18.0 g H2O (molar mass) 1 mol O2 = 6.02  1023 molecules O2 UNKNOWN molecules of oxygen = ? molecules O2 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.5 Calculate Solve for the unknown. 2 Start with the given quantity, and convert from mass to moles. Remember to also start your calculations with the given quantity, even if the given quantity is a product in the reaction. 1 mol H2O 18.0 g H2O Given quantity Change to moles 29.2 g H2O  Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.5 Calculate Solve for the unknown. 2 Then, convert from moles of reactant to moles of product. 1 mol O2  2 mol H2O 29.2 g H2O  1 mol H2O 18.0 g H2O Given quantity Change to moles Mole ratio Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.5 Calculate Solve for the unknown. 2 Finish by converting from moles to molecules. 6.02  1023 molecules O2 1 mol O2  2 mol H2O 29.2 g H2O  1 mol H2O 18.0 g H2O Given quantity Change to moles Mole ratio Change to molecules Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.5 Calculate Solve for the unknown. 2 Finish by converting from moles to molecules. 6.02  1023 molecules O2 1 mol O2  2 mol H2O 29.2 g H2O  1 mol H2O 18.0 g H2O Given quantity Change to moles Mole ratio Change to molecules = 4.88  1023 molecules O2 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.5 Evaluate Does the result make sense? 3 The given mass of water should produce a little less than 1 mol of oxygen, or a little less than Avogadro’s number of molecules. The answer should have three significant figures. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Volume-Volume Stoichiometric Calculations
Sample Problem 12.6 Volume-Volume Stoichiometric Calculations Nitrogen monoxide and oxygen gas combine to form the brown gas nitrogen dioxide, which contributes to photochemical smog. How many liters of nitrogen dioxide are produced when 34 L of oxygen react with an excess of nitrogen monoxide? Assume conditions are at STP. 2NO(g) + O2(g)  2NO2(g) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.6 Analyze List the knowns and the unknown. 1 For gaseous reactants and products at STP, 1 mol of a gas is equal to 22.4 L. KNOWNS volume of oxygen = 34 L O2 2 mol NO2/1 mol O2 (mole ratio in balanced equation) 1 mol O2 = 22.4 L O2 (at STP) 1 mol NO2 = 22.4 L NO2 (at STP) UNKNOWN volume of nitrogen dioxide = ? L NO2 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.6 Calculate Solve for the unknown. 2 Start with the given quantity, and convert from volume to moles by using the volume ratio. Did you notice that the 22.4 L/mol factors canceled out? This will always be true in a volume-volume problem. 34 L O2  1 mol O2 22.4 L O2 Given quantity Change to moles Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.6 Calculate Solve for the unknown. 2 Then, convert from moles of reactant to moles of product by using the correct mole ratio. 34 L O2   1 mol O2 22.4 L O2 2 mol NO2 Given quantity Change to moles Mole ratio Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.6 Calculate Solve for the unknown. 2 Finish by converting from moles to liters. Use the volume ratio. 34 L O2    1 mol O2 22.4 L O2 1 mol NO2 22.4 L NO2 2 mol NO2 Given quantity Change to moles Mole ratio Change to liters = 68 L NO2 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.6 Evaluate Does the result make sense? 3 Because 2 mol NO2 are produced for each 1 mol O2 that reacts, the volume of NO2 should be twice the given volume of O2. The answer should have two significant figures. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Finding the Volume of a Gas Needed for a Reaction
Sample Problem 12.7 Finding the Volume of a Gas Needed for a Reaction Assuming STP, how many milliliters of oxygen are needed to produce 20.4 mL SO3 according to this balanced equation? 2SO2(g) + O2(g)  2SO3(g) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.7 Analyze List the knowns and the unknown. 1 For a reaction involving gaseous reactants or products, the coefficients also indicate relative amounts of each gas. You can use the volume ratios in the same way you have used mole ratios. KNOWNS volume of sulfur trioxide = 20.4 mL 2 mL SO3/1 mL O2 (volume ratio in balanced equation) UNKNOWN volume of oxygen = ? mL O2 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.7 Calculate Solve for the unknown. 2 Multiply the given volume by the appropriate volume ratio. 20.4 mL SO3  = 10.2 mL O2 1 mL O2 2 mL SO3 The volume ratio can be written using milliliters as the units instead of liters. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.7 Evaluate Does the result make sense? 3 Because the volume ratio is 2 volumes SO3 to 1 volume O2, the volume of O2 should be half the volume of SO3. The answer should have three significant figures. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Methane burns in air by the following reaction: CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) What volume of water vapor is produced at STP by burning 501 g of methane? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Methane burns in air by the following reaction: CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) What volume of water vapor is produced at STP by burning 501 g of methane? 22.4 L H2O 1 mol H2O 501 g CH4  16.05 g CH4 1 mol CH4  2 mol H2O = 1.40  103 L H2O Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Key Concepts In chemical calculations, mole ratios are used to convert between moles of reactant and moles of product, between moles of reactants, or between moles of products. In a typical stoichiometric problem, the given quantity is first converted to moles. Then, the mole ratio from the balanced equation is used to calculate the moles of the wanted substance. Finally, the moles are converted to any other unit of measurement related to the unit mole. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

b mol W a mol G xb x mol G  = mol W a
Key Equation Mole-mole relationship for aG  bW: b mol W a mol G xb x mol G  = mol W a Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Glossary Terms mole ratio: a conversion factor derived from the coefficients of a balanced chemical equation interpreted in terms of moles Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

BIG IDEA The Mole and Quantifying Matter Mole ratios from the balanced equation are used to calculate the amount of a reactant or product in a chemical reaction from a given amount of one of the reactants or products. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Chapter 12 Stoichiometry 12.3 Limiting Reagent and Percent Yield
12.1 The Arithmetic of Equations 12.2 Chemical Calculations 12.3 Limiting Reagent and Percent Yield Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CHEMISTRY & YOU What determines how much product you can make? If a carpenter had two tabletops and seven table legs, he would have difficulty building more than one functional four-legged table. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Limiting and Excess Reagents
How is the amount of product in a reaction affected by an insufficient quantity of any of the reactants? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

To make tacos, you need enough meat, cheese, lettuce, tomatoes, sour cream, salsa, and seasonings. If you have only 2 taco shells, the quantity of taco shells will limit the number of tacos you can make. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

To make tacos, you need enough meat, cheese, lettuce, tomatoes, sour cream, salsa, and seasonings. If you have only 2 taco shells, the quantity of taco shells will limit the number of tacos you can make. Thus, the taco shells are the limiting ingredient. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

A balanced chemical equation is a chemist’s recipe. Chemical Equations N2(g) + 3H2(g) 2NH3(g) “Microscopic recipe” 1 molecule N2 3 molecules H2 2 molecules NH3 “Macroscopic recipe” 1 mol N2 3 mol H2 2 mol NH3 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

A balanced chemical equation is a chemist’s recipe. Chemical Equations N2(g) + 3H2(g) 2NH3(g) “Microscopic recipe” 1 molecule N2 3 molecules H2 2 molecules NH3 “Macroscopic recipe” 1 mol N2 3 mol H2 2 mol NH3 What would happen if two molecules (moles) of N2 reacted with three molecules (moles) of H2? Experimental Conditions Reactants Products Before reaction 2 molecules N molecules H2 0 molecules NH3 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Experimental Conditions Reactants Products Before reaction 2 molecules N molecules H2 0 molecules NH3 After reaction 1 molecule N molecules H2 2 molecules NH3 Before the reaction takes place, N2 and H2 are present in a 2:3 molecule (mole) ratio. As the reaction takes place, one molecule (mole) of N2 reacts with 3 molecules (moles) of H2 to produce two molecules (moles) of NH3. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Experimental Conditions Reactants Products Before reaction 2 molecules N molecules H2 0 molecules NH3 After reaction 1 molecule N molecules H2 2 molecules NH3 All the H2 has now been used up, and the reaction stops. One molecule (mole) of unreacted N2 is left in addition to the two molecules (moles) of NH3 that have been produced by the reaction. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Experimental Conditions Reactants Products Before reaction 2 molecules N molecules H2 0 molecules NH3 After reaction 1 molecule N molecules H2 2 molecules NH3 In this reaction, only the hydrogen is completely used up. H2 is the limiting reagent, or the reactant that determines the amount of product that can be formed by a reaction. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Experimental Conditions Reactants Products Before reaction 2 molecules N molecules H2 0 molecules NH3 After reaction 1 molecule N molecules H2 2 molecules NH3 The reactant that is not completely used up in a reaction is called the excess reagent. In this example, nitrogen is the excess reagent because some nitrogen remains unreacted. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

What determines how much product you can make in a chemical reaction?
CHEMISTRY & YOU What determines how much product you can make in a chemical reaction? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CHEMISTRY & YOU What determines how much product you can make in a chemical reaction? A limited quantity of any of the reactants that are needed to make a product will limit the amount of product that forms. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Determining the Limiting Reagent in a Reaction
Sample Problem 12.8 Determining the Limiting Reagent in a Reaction Copper reacts with sulfur to form copper(I) sulfide according to the following balanced equation: 2Cu(s) + S(s)  Cu2S(s) What is the limiting reagent when 80.0 g Cu reacts with 25.0 g S? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

1 KNOWNS UNKNOWN Analyze List the knowns and the unknown.
Sample Problem 12.8 Analyze List the knowns and the unknown. 1 The number of moles of each reactant must first be found. The balanced equation is used to calculate the number of moles of one reactant needed to react with the given amount of the other reactant. KNOWNS UNKNOWN mass of copper = 80.0 g Cu mass of sulfur = 25.0 g S molar mass of Cu = 63.5 g/mol molar mass of S = 32.1 g/mol 1 mol S/2 mol Cu limiting reagent = ? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Start with one of the reactants and convert from mass to moles.
Sample Problem 12.8 Calculate Solve for the unknown. 2 Start with one of the reactants and convert from mass to moles. 80.0 g Cu  = 1.26 mol Cu 63.5 g Cu 1 mol Cu Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Then, convert the mass of the other reactant to moles.
Sample Problem 12.8 Calculate Solve for the unknown. 2 Then, convert the mass of the other reactant to moles. 25.0 g S  = mol S 32.1 g S 1 mol S Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.8 Calculate Solve for the unknown. 2 Now, convert moles of Cu to moles of S needed to react with 1.26 moles of Cu. 1.26 mol Cu = mol S 2 mol Cu 1 mol S Given quantity Mole ratio Needed amount Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Compare the amount of sulfur needed with the given amount of sulfur.
Sample Problem 12.8 Calculate Solve for the unknown. 2 Compare the amount of sulfur needed with the given amount of sulfur. 0.630 mol S (amount needed to react) <0.779 mol S (given amount) Sulfur is in excess, so copper is the limiting reagent. It doesn’t matter which reactant you use. If you used the actual amount of moles of S to find the amount of copper needed, then you would still identify copper as the limiting reagent. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

3 Evaluate Do the results make sense?
Sample Problem 12.8 Evaluate Do the results make sense? 3 Since the ratio of the given mol Cu to mol S was less than the ratio (2:1) from the balanced equation, copper should be the limiting reagent. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Using Limiting Reagent to Find the Quantity of a Product
Sample Problem 12.9 Using Limiting Reagent to Find the Quantity of a Product What is the maximum number of grams of Cu2S that can be formed when 80.0 g Cu reacts with 25.0 g S? 2Cu(s) + S(s)  Cu2S(s) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.9 Analyze List the knowns and the unknown. 1 The limiting reagent, which was determined in the previous sample problem, is used to calculate the maximum amount of Cu2S formed. KNOWNS limiting reagent = 1.26 mol Cu (from sample problem 12.8) 1 mol Cu2S = g Cu2S (molar mass) 1 mol Cu2S/2 mol Cu (mole ratio from balanced equation) UNKNOWN Yield = ? g Cu2S Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.9 Calculate Solve for the unknown. 2 Start with the moles of the limiting reagent and convert to moles of the product. Use the mole ratio from the balanced equation. 1.26 mol Cu  2 mol Cu 1 mol Cu2S Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Finish the calculation by converting from moles to mass of product.
Sample Problem 12.9 Calculate Solve for the unknown. 2 Finish the calculation by converting from moles to mass of product. 159.1 g Cu2S 1 mol Cu2S 1.26 mol Cu  2 mol Cu  = 1.00  102 g Cu2S Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

3 Evaluate Do the results make sense?
Sample Problem 12.9 Evaluate Do the results make sense? 3 Copper is the limiting reagent in this reaction. The maximum number of grams of Cu2S produced should be more than the amount of copper that initially reacted because copper is combining with sulfur. However, the mass of Cu2S produced should be less than the total mass of the reactants (105.0 g) because sulfur was in excess. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Rust forms when iron, oxygen, and water react
Rust forms when iron, oxygen, and water react. One chemical equation for the formation of rust is 2Fe + O2 + 2H2O  2Fe(OH)2 If 7.0 g of iron and 9.0 g of water are available to react, which is the limiting reagent? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Rust forms when iron, oxygen, and water react
Rust forms when iron, oxygen, and water react. One chemical equation for the formation of rust is 2Fe + O2 + 2H2O  2Fe(OH)2 If 7.0 g of iron and 9.0 g of water are available to react, which is the limiting reagent? 7.00 g Fe    = 2.26 g H20 1 mol Fe 55.85 g Fe 18.0 g H2O 1 mol H2O 2 mol Fe 2 mol H2O Only 2.26 g H2O are needed to react with 7.0 g Fe. Therefore, Fe is the limiting reagent. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Percent Yield When a balanced chemical equation is used to calculate the amount of product that will form during a reaction, the calculated value represents the theoretical yield. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Percent Yield When a balanced chemical equation is used to calculate the amount of product that will form during a reaction, the calculated value represents the theoretical yield. The theoretical yield is the maximum amount of product that could be formed from given amounts of reactants. The amount of product that actually forms when the reaction is carried out in the laboratory is called the actual yield. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Percent Yield The percent yield is the ratio of the actual yield to the theoretical yield expressed as a percent. percent yield = actual yield theoretical yield  100% Because the actual yield of a chemical reaction is often less than the theoretical yield, the percent yield is usually less than 100%. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Percent Yield The percent yield is a measure of the efficiency of a reaction carried out in the laboratory. The mass of the reactant is measured. The mass of one of the products, the actual yield, is measured. The percent yield is calculated. The reactant is heated. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Many factors cause percent yields to be less than 100%.
Reactions do not always go to completion; when a reaction is incomplete, less than the calculated amount of product is formed. Impure reactants and competing side reactions may cause unwanted products to form. Actual yield can be lower than the theoretical yield due to a loss of product during filtration or in transferring between containers. If reactants or products have not been carefully measured, a percent yield of 100% is unlikely. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CaCO3(s)  CaO(s) + CO2(g)
Sample Problem 12.10 Calculating the Theoretical Yield of a Reaction Calcium carbonate, which is found in seashells, is decomposed by heating. The balanced equation for this reaction is CaCO3(s)  CaO(s) + CO2(g) D What is the theoretical yield of CaO if 24.8 g CaCO3 is heated? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.10 Analyze List the knowns and the unknown. 1 Calculate the theoretical yield using the mass of the reactant. KNOWNS mass of CaCO3 = 24.8 g CaCO3 1 mol CaCO3 = g CaCO3 (molar mass) 1 mol CaO = 56.1 g CaO (molar mass) 1 mol CaO/1 mol CaCO3 (mole ratio from balanced equation) UNKNOWN theoretical yield = ? g CaO Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.10 Calculate Solve for the unknown. 2 Start with the mass of the reactant and convert to moles of the reactant. 24.8 g CaCO3  100.1 g CaCO3 1 mol CaCO3 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Next, convert to moles of the product using the mole ratio.
Sample Problem 12.10 Calculate Solve for the unknown. 2 Next, convert to moles of the product using the mole ratio. 24.8 g CaCO3   100.1 g CaCO3 1 mol CaCO3 1 mol CaO Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Finish by converting from moles to mass of the product.
Sample Problem 12.10 Calculate Solve for the unknown. 2 Finish by converting from moles to mass of the product. 24.8 g CaCO3    100.1 g CaCO3 1 mol CaCO3 1 mol CaO 56.1 g CaO If there is an excess of a reactant, then there is more than enough of that reactant and it will not limit the yield of the reaction. = 13.9 g CaO Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

3 Evaluate Does the result make sense?
Sample Problem 12.10 Evaluate Does the result make sense? 3 The mole ratio of CaO to CaCO3 is 1:1. The ratio of their masses in the reaction should be the same as the ratio of their molar masses, which is slightly greater than 1:2. The result of the calculations shows that the mass of CaO is slightly greater than half the mass of CaCO3. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CaCO3(s)  CaO(s) + CO2(g)
Sample Problem 12.11 Calculating the Percent Yield of a Reaction What is the percent yield if 13.1 g CaO is actually produced when 24.8 g CaCO3 is heated? Calculate the theoretical yield first. Then you can calculate the percent yield. CaCO3(s)  CaO(s) + CO2(g) D Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

1 Analyze List the knowns and the unknown.
Sample Problem 12.11 Analyze List the knowns and the unknown. 1 Use the equation for percent yield. The theoretical yield for this problem was calculated in Sample Problem KNOWNS actual yield = 13.1 g CaO theoretical yield = 13.9 g CaO (from sample problem 12.10) percent yield = actual yield theoretical yield  100% UNKNOWN percent yield = ? % Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.11 Calculate Solve for the unknown. 2 Substitute the values for actual yield and theoretical yield into the equation for percent yield. percent yield =  100% = 94.2% 13.1 g CaO 13.9 g CaO Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

3 Evaluate Does the result make sense?
Sample Problem 12.11 Evaluate Does the result make sense? 3 In this example, the actual yield is slightly less than the theoretical yield. Therefore, the percent yield should be slightly less than 100%. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Carbon tetrachloride, CCl4, is a solvent that was once used in large amounts in dry cleaning. One reaction that produces carbon tetrachloride is CS2 + 3Cl2  CCl4 + S2Cl2 What is the percent yield of CCl4 if 617 kg is produced from the reaction of 312 kg of CS2? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

What is the percent yield of CCl4 if 617 kg is produced from the reaction of 312 kg of CS2?
CS2 + 3Cl2  CCl4 + S2Cl2 3.12  105 g CS2    g CS2 1 mol CS2 1 mol CCl4 g CCl4 = 6.30  105 g CCl4 = 630 kg CCl4 Percent yield =  100% = 97.9% 617 kg CCl4 630 kg CCl4 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Key Concepts and Key Equation
In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms. The percent yield is a measure of the efficiency of a reaction performed in the laboratory. percent yield = actual yield theoretical yield  100% Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Glossary Terms limiting reagent: any reactant that is used up first in a chemical reaction; it determines the amount of product that can be formed in the reaction excess reagent: a reagent present in a quantity that is more than sufficient to react with a limiting reagent; any reactant that remains after the limiting reagent is used up in a chemical reaction Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Glossary Terms theoretical yield: the amount of product that could form during a reaction calculated from a balanced chemical equation; it represents the maximum amount of product that could be formed from a given amount of reactant actual yield: the amount of product that forms when a reaction is carried out in the laboratory percent yield: the ratio of the actual yield to the theoretical yield for a chemical reaction expressed as a percentage; a measure of the efficiency of a reaction Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

BIG IDEA The Mole and Quantifying Matter The percent yield of a reaction can be calculated from the actual yield and theoretical yield of the reaction. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Chapter 12 Stoichiometry 12.1 The Arithmetic of Equations

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Chapter 12 Stoichiometry 12.1 The Arithmetic of Equations

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