* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download 258-261
Biochemistry wikipedia , lookup
Electrolysis of water wikipedia , lookup
Chemical equilibrium wikipedia , lookup
Process chemistry wikipedia , lookup
Lewis acid catalysis wikipedia , lookup
Double layer forces wikipedia , lookup
Chemical reaction wikipedia , lookup
Dimensional analysis wikipedia , lookup
Electrochemistry wikipedia , lookup
Computational chemistry wikipedia , lookup
Click chemistry wikipedia , lookup
Physical organic chemistry wikipedia , lookup
Rate equation wikipedia , lookup
Relativistic quantum mechanics wikipedia , lookup
Chemical thermodynamics wikipedia , lookup
Transition state theory wikipedia , lookup
Organosulfur compounds wikipedia , lookup
Atomic theory wikipedia , lookup
Gas chromatography–mass spectrometry wikipedia , lookup
Page 1 of 4 (continued) for converting moles of H2 to moles of NH3: 2 mol NH3 1.30 E mol H2 0.867 mol NH3 3E mol H2 Check: Notice that the number of moles of NH3 that can be produced is less than 1.30. This answer makes sense because it takes 3 moles of H2 to yield 2 moles of NH3. Thus the number of moles of NH3 produced (0.867) should be smaller than the number of moles of H2 consumed (1.30). Focus Questions Sections 9.1–9.2 1. What part of a chemical equation allows us to determine how much product we can get from a known amount of reactants? 2. Why is it important to balance a chemical equation before answering any questions about it? 3. Give at least three different ways to interpret the coefficient in a chemical equation. 4. How does a mole ratio help to determine the amount of product produced from a known number of moles of reactant? 5. What number of moles of O2(g) is required to react with 3.6 mol SO2(g) in the following reaction? 2SO2(g) O2(g) n 2SO3(g) Mass Calculations Objective: To learn to relate masses of reactants and products in a chemical reaction. I n the last section we saw how to use the balanced equation for a reaction to calculate the numbers of moles of reactants and products for a particular case. However, moles represent numbers of molecules, and we cannot count molecules directly. In chemistry we count by weighing. Therefore, in this section we will review the procedures for converting between moles and masses and will see how these procedures are applied to chemical calculations. To develop these procedures we will consider the combustion of propane, C3H8. Propane, when used as a fuel, reacts with oxygen to produce carbon dioxide and water according to the following unbalanced equation: C3H8(g) O2(g) n CO2(g) H2O(g) PROBLEM SOLVING Always balance the equation for the reaction first. 258 What mass of oxygen will be required to react exactly with 44.1 g of propane? To deal with the amounts of reactants and products, we first need the balanced equation for this reaction: C3H8(g) 5O2(g) n 3CO2(g) 4H2O(g) Chapter 9 Chemical Quantities Page 2 of 4 Next, let’s summarize what we know and what we want to find. W H AT I F ? Your lab partner has made the observation that we always measure the mass of chemicals but then use mole ratios to balance equations. What We Know: ● The balanced equation for the reaction ● The mass of propane available (44.1 g) What We Want to Calculate: What if your lab partner decided to balance equations by using masses as coefficients? Is this possible? Why or why not? ● The mass of oxygen (O2) required to react exactly with all the propane Our problem, in schematic form, is 96.1 g propane ? grams of O2 requires Our overall plan of attack is as follows: 1. We are given the number of grams of propane, so we must convert to moles of propane (C3H8), because the balanced equation deals in moles rather than grams. 2. Next, we can use the coefficients in the balanced equation to determine the moles of oxygen (O2) required. 3. Finally, we will use the molar mass of O2 to calculate grams of oxygen. We can sketch this strategy as follows: C3H8(g) 5O2(g) 44.1 g C3H8 ? grams of O2 1 3 ? moles of C3H8 1 2 n 3CO2(g) 4H2O(g) ? moles of O2 Thus the first question we must answer is, How many moles of propane are present in 44.1 g of propane? The molar mass of propane is 44.09 g (3 12.01 8 1.008). The moles of propane present can be calculated as follows: 1 mol C3H8 44.1 g C3H8 1.00 mol C3H8 44.09 g C3H8 2 Next we recognize that each mole of propane reacts with 5 mol of oxygen. This gives us the equivalence statement 1 mol C3H8 5 mol O2 from which we construct the mole ratio 5 mol O2 1 mol C3H8 that we need to convert from moles of propane molecules to moles of oxygen molecules. 5 mol O2 1.00 mol C3H8 5.00 mol O2 1 mol C3H8 9.3 Mass Calculations 259 Page 3 of 4 3 Notice that the mole ratio is set up so that the moles of C3H8 cancel and the resulting units are moles of O2. Because the original question asked for the mass of oxygen needed to react with 44.1 g of propane, we must convert the 5.00 mol of O2 to grams, using the molar mass of O2 (32.00 2 16.00). 32.0 g O2 5.00 mol O2 160. g O2 1 mol O2 Therefore, 160 g of oxygen is required to burn 44.1 g of propane. We can summarize this problem by writing out a “conversion string” that shows how the problem was done. 1 3 2 5 mol O2 1 mol C3H8 32.0 g O2 160. g O2 44.1 g C3H8 1 mol C3H8 44.09 g C3H8 1 mol O2 PROBLEM SOLVING Use units as a check to see that you have used the correct conversion factors (mole ratios). This is a convenient way to make sure the final units are correct. The procedure we have followed is summarized below. C3H8(g) 5O2(g) 44.1 g C3H8 160. g O2 1 3 Use molar mass of C3H8 (44.09 g) Use molar mass of O2 (32.0 g) 1 3 1.00 mol C3H8 2 Use mole ratio: 5 mol O2 1 mol C3H8 2 n 3CO2(g) 4H2O(g) 5.00 mol O2 This example illustrates in detail how to deal with masses of reactants in chemical equations. To see whether these ideas are clear to you, try the following Self-Check Exercises. Self-Check Exercise 9.2 What mass of carbon dioxide is produced when 44.1 g of propane reacts with sufficient oxygen? Self-Check Exercise 9.3 Calculate the mass of water formed by the complete reaction of 44.1 g of propane with oxygen. To illustrate how to deal with masses in a chemical equation, consider the following example. 260 Chapter 9 Chemical Quantities Page 4 of 4 Example 9.5 Mass Relationships in Chemical Reactions Consider the reaction of powdered aluminum metal and finely ground iodine to produce aluminum iodide. The balanced equation for this vigorous chemical reaction is 2Al(s) 3I2(s) n 2AlI3(s) Calculate the mass of I2(s) needed to just react with 35.0 g Al(s). Solution Let’s summarize what we know and what we need to calculate. What We Know: ● The balanced equation for the reaction ● The mass of Al(s) available What We Want to Calculate: ● The mass of iodine (I2) required to react with all of the aluminum (35.0 g) Let’s summarize our plan: Aluminum (left) and iodine (right) shown at the top, react vigorously to form aluminum iodide. The purple cloud results from excess iodine vaporized by the heat of the reaction. 1. We first need to convert from 35.0 g aluminum to moles of aluminum because the balanced equation deals in moles rather than grams. 2. We will use the balanced equation to find the moles of I2 required to react with all of the aluminum. 3. Once we know the moles of I2(s) required, we will use the molar mass for iodine to calculate the mass of I2(s) required. A schematic for this strategy is 2Al(s) 3I2(s) 35.0 g Al ? grams I2 ? moles of Al ? moles of I2 n 2AlI3(s) Using the molar mass of aluminum (26.98 g), we will first calculate the moles of aluminum present in 35.0 g Al: 1 mol Al 35.0 g Al 1.30 mol Al 26.98 g Al From the balanced equation for the reaction we see that 2 Al requires 3 I2 (continued) 9.3 Mass Calculations 261