Sample Exercise 1 Interpreting and Balancing Chemical Equations Solution (a)



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  • Solution
  • (a) The left box, which represents the reactants, contains two kinds of molecules, those composed of two oxygen atoms (O2) and those composed of one nitrogen atom and one oxygen atom (NO). The right box, which represents the products, contains only molecules composed of one nitrogen atom and two oxygen atoms (NO2).
  • (b) The unbalanced chemical equation is
  • O2 + NO → NO2 (unbalanced)
  • This equation has three O atoms on the left side of the arrow and two O atoms on the right side. We can increase the number of O atoms by placing a coefficient 2 on the product side:
  • O2 + NO → 2 NO2 (unbalanced)
  • The following diagram represents a chemical reaction in which the red spheres are oxygen atoms and the blue spheres are nitrogen atoms. (a) Write the chemical formulas for the reactants and products. (b) Write a balanced equation for the reaction. (c) Is the diagram consistent with the law of conservation of mass?
  • Sample Exercise 2.1 Interpreting and Balancing Chemical Equations
  • Solution (continued)
  • Now there are two N atoms and four O atoms on the right. Placing the coefficient 2 in front of NO balances both the number of N atoms and O atoms:
  • O2 + 2 NO → 2 NO2 (balanced)
  • (c) The left box (reactants) contains four O2 molecules and eight NO molecules. Thus, the molecular ratio is one O2 for each two NO as required by the balanced equation. The right box (products) contains eight NO2 molecules. The number of NO2 molecules on the right equals the number of NO molecules on the left as the balanced equation requires. Counting the atoms, we find eight N atoms in the eight NO molecules in the box on the left. There are also 4  2 = 8 O atoms in the O2 molecules and eight O atoms in the NO molecules, giving a total of 16 O atoms. In the box on the right, we find eight N atoms and 8  2 = 16 O atoms in the eight NO2 molecules. Because there are equal numbers of both N and O atoms in the two boxes, the drawing is consistent with the law of conservation of mass.
  • In the following diagram, the white spheres represent hydrogen atoms, and the blue spheres represent nitrogen atoms. To be consistent with the law of conservation of mass, how many NH3 molecules should be shown in the right box?
  • Answer: Six NH3 molecules.
  • Sample Exercise 2.2 Balancing Chemical Equations
  • Balance this equation:
  • Na(s) + H2O(l) → NaOH(aq) + H2(g)
  • Solution
  • Begin by counting each kind of atom on both sides of the arrow. The Na and O atoms are balanced—one Na and one O on each side—but there are two H atoms on the left and three H atoms on the right. Thus, we need to increase the number of H atoms on the left. To begin balancing H, let’s try placing the coefficient 2 in front of H2O:
  • Beginning this way does not balance H but does increase the number of H atoms among the reactants, which we need to do. Adding the coefficient 2 causes O to be unbalanced; we will take care of that after we balance H. Now that we have 2 H2O on the left, we can balance H by
  • putting the coefficient 2 in front of NaOH on the right:
  • Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g)
  • Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g)
  • Sample Exercise 2.2 Balancing Chemical Equations
  • Balance the following equations by providing the missing coefficients:
  • (a) _Fe(s) + _O2(g) → _Fe2O3(s)
  • (b) _C2H4(g) + _O2(g) → _CO2(g) + _H2O(g)
  • (c) _Al(s) + _HCl(aq) → _AlCl3(aq) + _H2(g)
  • Practice Exercise
  • Solution (continued)
  • Balancing H in this way fortuitously brings O into balance. But notice that Na is now unbalanced, with one Na on the left and two on the right. To rebalance Na, we put the coefficient 2 in front of the reactant:
  • 2Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g)
  • Finally, we check the number of atoms of each element and find that we have two Na atoms, four H atoms, and two O atoms on each side of the equation. The equation is balanced.
  • Comment Notice that in balancing this equation, we moved back and forth placing a coefficient in front of H2O, then NaOH, and finally Na. In balancing equations, we often find ourselves following this pattern of moving back and forth from one side of the arrow to the other, placing coefficients first in front of a formula on one side and then in front of a formula on the other side until the equation is balanced. You can always tell if you have balanced your equation correctly, no matter how you did it, by checking that the number of atoms of each element is the same on both sides of the arrow.
  • Answers: (a) 4, 3, 2; (b) 1, 3, 2, 2; (c) 2, 6, 2, 3
  • Sample Exercise 2.3 Writing Balanced Equations for Combination and
  • Decomposition Reactions
  • Solution
  • (a) The symbol for lithium is Li. With the exception of mercury, all metals are solids at room temperature. Fluorine occurs as a diatomic molecule (see Figure 2.19). Thus, the reactants are Li(s) and F2(g). The product will be composed of a metal and a nonmetal, so we expect it to be an ionic solid. Lithium ions have a 1+ charge, Li+, whereas fluoride ions have a 1– charge, F–.Thus, the chemical formula for the product
  • is LiF. The balanced chemical equation is
  • 2 Li(s) + F2(g) → 2 LiF(s)
  • (b) The chemical formula for barium carbonate is BaCO3. As noted in the text, many
  • metal carbonates decompose to form metal oxides and carbon dioxide when heated.
  • In Equation 3.7, for example, CaCO3 decomposes to form CaO and CO2. Thus, we
  • would expect that BaCO3 decomposes to form BaO and CO2. Barium and calcium are
  • both in group 2A in the periodic table, which further suggests they would react in the
  • same way:
  • BaCO3(s) → BaO(s) + CO2(g)
  • Write balanced equations for the following reactions: (a) The combination reaction that occurs when lithium metal and fluorine gas react. (b) The decomposition reaction that occurs when solid barium carbonate is heated. (Two products form: a solid and a gas.)
  • Sample Exercise 2.3 Writing Balanced Equations for Combination and
  • Decomposition Reactions
  • Write balanced chemical equations for the following reactions: (a) Solid mercury(II) sulfide decomposes into its component elements when heated. (b) The surface of aluminum metal undergoes a combination reaction with oxygen in the air.
  • Answer: (a) HgS(s) → Hg(l) + S(s) (b) 4 Al(s) + 3 O2(g) → 2 Al2O3(s)
  • Practice Exercise
  • Sample Exercise 2.4 Writing Equations for Combustion Reactions
  • Write the balanced equation for the reaction that occurs when methanol, CH3OH(l), is burned in air.
  • Solution
  • When any compound containing C, H, and O is combusted, it reacts with the O2(g) in air to produce CO2(g) and H2O(g). Thus, the unbalanced equation is
  • CH3OH(l) + O2(g) → CO2(g) + H2O(g)
  • In this equation the C atoms are balanced with one carbon on each side of the arrow. Because CH3OH has four H atoms, we place the coefficient 2 in front of H2O to balance the H atoms:
  • CH3OH(l) + O2(g) → CO2(g) + 2 H2O(g)
  • Adding the coefficient balances H but gives four O atoms in the products. Because there are only three O atoms in the reactants (one in CH3OH and two in O2), we are not finished yet. We can place the fractional coefficient 2/3 in front of O2 to give a total of four O atoms in the reactants
  • (there are 2/3  2 = 3 O atoms in 3/2 O2):
  • Although the equation is now balanced, it is not in its most conventional form because
  • it contains a fractional coefficient. If we multiply each side of the equation by 2,
  • we will remove the fraction and achieve the following balanced equation:
  • 2 CH3OH(l) + 3 O2(g) → 2 CO2(g) + 4 H2O(g)
  • Sample Exercise 2.4 Writing Equations for Combustion Reactions
  • Write the balanced equation for the reaction that occurs when ethanol, C2H5OH(l), is
  • burned in air.
  • Answer: C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g)
  • Practice Exercise
  • Sample Exercise 2.5 Calculating formula Weights
  • Calculate the formula weight of (a) sucrose, C12H22O11 (table sugar), and (b) calcium nitrate, Ca(NO3)2.
  • Calculate the formula weight of (a) Al(OH)3 and (b) CH3OH.
  • Practice Exercise
  • Solution
  • (a) By adding the atomic weights of the atoms in sucrose, we find the formula weight to be 342.0 amu:
  • (b) If a chemical formula has parentheses, the subscript outside the parentheses is a multiplier for all atoms inside.
  • Thus, for Ca(NO3)2, we have
  • Answer: (a) 78.0 amu, (b) 32.0 amu
  • Sample Exercise 2.6 Calculating Percentage Composition
  • Calculate the percentage of nitrogen, by mass, in Ca(NO3)2.
  • Practice Exercise
  • Calculate the percentage of carbon, hydrogen, and oxygen (by mass) in C12H22O11.
  • Solution
  • Let’s examine this question using the problem-solving steps in the “Strategies in Chemistry: Problem Solving” essay that appears on the next page.
  • Analyze We are given a chemical formula, C12H22O11, and asked to calculate the percentage by mass of its component elements (C, H, and O).
  • Plan We can use Equation 3.10, relying on a periodic table to obtain the atomic weight of each component element. The atomic weights are first used to determine the formula weight of the compound. (The formula weight of C12H22O11, 342.0 amu, was calculated in Sample Exercise 3.5.) We must then do three calculations, one for each element.
  • Solve Using Equation 3.10, we have
  • Check The percentages of the individual elements must add up to 100%, which they do in this case. We could have used more significant figures for our atomic weights, giving more significant figures for our percentage composition, but we have adhered to our suggested guideline of rounding atomic weights to one digit beyond the decimal point.
  • Answer: 17.1%
  • Sample Exercise 2.7 Estimating Numbers in Atoms
  • Without using a calculator, arrange the following samples in order of increasing number of O atoms: 1 mol H2O, 1 mol CO2, 3  1023 molecules O3.
  • Practice Exercise
  • Solution
  • Analyze We are given amounts of different substances expressed in grams, moles, and number of molecules and asked to arrange the samples in order of increasing numbers of C atoms.
  • Plan To determine the number of C atoms in each sample, we must convert g 12C, 1 mol C2H2, and 9  1023 molecules CO2 all to numbers of C atoms. To make these conversions, we use the definition of mole and Avogadro’s number.
  • Solve A mole is defined as the amount of matter that contains as many units of the matter as there are C atoms in exactly 12 g of 12C. Thus, 12 g of 12C contains 1 mol of C atoms (that is, 6.02  1023 C atoms). One mol of C2H2 contains 6  1023 C2H2 molecules. Because there are two C atoms in each C2H2 molecule, this sample contains 12  1023 C atoms. Because each CO2 molecule contains one C atom, the sample of CO2 contains 9  1023 C atoms. Hence, the order is 12 g 12C (6  1023 C atoms) < 9  1023 CO2 molecules (9  1023 C atoms) < 1 mol C2H2 (12  1023 C atoms).
  • Check We can check our results by comparing the number of moles of C atoms in each sample because the number of moles is proportional to the number of atoms. Thus, 12 g of 12C is 1 mol C; 1 mol of C2H2 contains 2 mol C, and 9  1023 molecules of CO2 contain 1.5 mol C, giving the same order as above: 12 g 12C (1 mol C) < 9  1023 CO2 molecules (1.5 mol C) < 1 mol C2H2 (2 mol C).
  • Without using a calculator, arrange the following samples in order of increasing numbers of carbon atoms: 12 g 12C, 1 mol C2H2, 9  1023 molecules of CO2.
  • Answer: 1 mol H2O (6  1023 O atoms) 3  1023 molecules O3 (9  1023 O atoms) 1 mol CO2 (12  1023 O atoms)
  • Calculate the number of H atoms in 0.350 mol of C6H12O6.
  • Solution
  • Analyze We are given both the amount of a substance (0.350 mol) and its chemical formula (C6H12O6). The unknown is the number of H atoms in the sample.
  • Plan Avogadro’s number provides the conversion factor between the number of moles of C6H12O6 and the number of molecules of C6H12O6. Once we know the number of molecules of C6H12O6, we can use the chemical formula, which tells us that each molecule of C6H12O6 contains 12 H atoms. Thus, we convert moles of C6H12O6 to molecules of C6H12O6 and then determine the number of atoms of H from the number of molecules of C6H12O6:
  • Moles C6H12O6 → molecules C6H12O6 → atoms H
  • Check The magnitude of our answer is reasonable. It is a large number about the magnitude of Avogadro’s number. We can also make the following ballpark calculation: Multiplying 0.35  6  1023 gives about
  • 2  1023 molecules. Multiplying this result by 12 gives 24  1023 = 2.4  1024 H atoms, which agrees with the previous, more detailed calculation. Because we were asked for the number of H atoms, the units of our answer are correct. The given data had three significant figures, so our answer has three significant figures.
  • Sample Exercise 2.8 Converting Moles to Atoms
  • How many oxygen atoms are in (a) 0.25 mol Ca(NO3)2 and (b) 1.50 mol of sodium carbonate?
  • Answer: (a) 9.0  1023, (b) 2.71  1024
  • Practice Exercise
  • Calculate the number of H atoms in 0.350 mol of C6H12O6.
  • Sample Exercise 2.9 Calculating Molar Mass
  • Calculate the molar mass of Ca(NO3)2.
  • Practice Exercise
  • What is the mass in grams of 1.000 mol of glucose, C6H12O6?
  • Solution
  • Analyze We are given a chemical formula and asked to determine its molar mass.
  • Plan The molar mass of a substance is found by adding the atomic weights of its component atoms.
  • Solve
  • Because glucose has a formula weight of 180.0 amu, one mole of this substance has a mass of 180.0 g. In other words, C6H12O6 has a molar mass of 180.0 g/mol.
  • Check The magnitude of our answer seems reasonable, and g/mol is the appropriate unit for the molar mass.
  • Comment Glucose is sometimes called dextrose. Also known as blood sugar, glucose is found widely in nature, occurring in honey and fruits. Other types of sugars used as food are converted into glucose in the stomach or liver before the body uses them as energy sources. Because glucose requires no conversion, it is often given intravenously to patients who need immediate nourishment. People who have diabetes must carefully monitor the amount of glucose in their blood (See “Chemistry and Life” box in Section 3.6).
  • Answer: 164.1 g/mol
  • Sample Exercise 2.10 Converting Grams to Moles
  • How many moles of sodium bicarbonate (NaHCO3) are in 508 g of NaHCO3?
  • Practice Exercise
  • Calculate the number of moles of glucose (C6H12O6) in 5.380 g of C6H12O6.
  • Solution
  • Analyze We are given the number of grams of a substance and its chemical formula and asked to calculate the number of moles.
  • Plan The molar mass of a substance provides the factor for converting grams to moles. The molar mass of C6H12O6 is 180.0 g/mol (Sample Exercise 3.9).
  • Solve Using 1 mol C6H12O6=180.0 g C6H12O6 to write the appropriate conversion factor, we have
  • Check Because 5.380 g is less than the molar mass, a reasonable answer is less than one mole. The units of our answer (mol) are appropriate. The original data had four significant figures, so our answer has four significant figures.
  • Answer: 6.05 mol NaHCO3
  • Sample Exercise 2.11 Converting Moles to Grams
  • What is the mass, in grams, of (a) 6.33 mol of NaHCO3 and (b) 3.0  10–5 mol of sulfuric acid?
  • Practice Exercise
  • Calculate the mass, in grams, of 0.433 mol of calcium nitrate.
  • Solution
  • Analyze We are given the number of moles and the name of a substance and asked to calculate the number of grams in the sample.
  • Plan To convert moles to grams, we need the molar mass, which we can calculate using the chemical formula and atomic weights.
  • Solve Because the calcium ion is Ca2+ and the nitrate ion is NO3–, calcium nitrate is Ca(NO3)2. Adding the atomic weights of the elements in the compound gives a formula weight of 164.1 amu. Using 1 mol Ca(NO3)2 = 164.1 g Ca(NO3)2 to write the appropriate conversion factor, we have
  • Check The number of moles is less than 1, so the number of grams must be less than the molar mass, 164.1 g. Using rounded numbers to estimate, we have 0.5  150 = 75 g. The magnitude of our answer is reasonable. Both the units (g) and the number of significant figures (3) are correct.
  • Answer: (a) 532 g, (b) 2.9  10–3 g
  • Sample Exercise 3.12 Calculating the Number of Molecules and Number
  • of Atoms from Mass
  • (a) How many glucose molecules are in 5.23 g of C6H12O6? (b) How many oxygen atoms are in this sample?
  • Solution
  • Analyze We are given the number of grams and the chemical formula and asked to calculate (a) the number of molecules and (b) the number of O atoms in the sample.
  • (a) Plan The strategy for determining the number of molecules in a given quantity of a substance is summarized in Figure 3.10. We must convert 5.23 g C6H12O6 to moles C6H12O6, which can then be converted to molecules C6H12O6. The first conversion uses the molar mass of C6H12O6:
  • 1 mol C6H12O6 = 180.0 g C6H12O6. The second conversion uses Avogadro’s number.
  • Solve
  • Molecules C6H12O6
  • Check The magnitude of the answer is reasonable. Because the mass we began with is less than a mole, there should be fewer than 6.02  1023 molecules. We can make a ballpark estimate of the answer: 5/200 = 2.5 10–2 mol; 2.5  10–2  6  1023 = 15  1021 = 1.5  1022 molecules. The units (molecules) and significant figures (three) are appropriate.
  • Sample Exercise 3.12 Calculating the Number of Molecules and Number
  • of Atoms from Mass
  • (a) How many nitric acid molecules are in 4.20 g of HNO3? (b) How many O atoms are in this sample?
  • Practice Exercise
  • Solution (continued)
  • (b) Plan To determine the number of O atoms, we use the fact that there are six O atoms in each molecule of C6H12O6. Thus, multiplying the number of molecules C6H12O6 by the factor (6 atoms O/1 molecule C6H12O6) gives the number of O atoms.
  • Solve
  • Check The answer is simply 6 times as large as the answer to part (a). The number of significant figures (three) and the units (atoms O) are correct.
  • Answer: (a) 4.01  1022 molecules HNO3, (b) 1.20  1023 atoms O
  • Sample Exercise 2.13 Calculating Empirical Formula
  • Ascorbic acid (vitamin C) contains 40.92% C, 4.58% H, and 54.50% O by mass. What is the empirical formula of ascorbic acid?
  • Solution
  • Solve We first assume, for simplicity, that we have exactly 100 g of material (although any mass can be used). In 100 g of ascorbic acid, therefore, we have
  • Second, we calculate the number of moles of each element:
  • Third, we determine the simplest whole-number ratio of moles by dividing each number of moles by the smallest number of moles, 3.406:
  • Analyze We are to determine an empirical formula of a compound from the mass percentages of its elements.
  • Plan The strategy for determining the empirical formula involves the three steps given in Figure 3.11.
  • Sample Exercise 2.13 Calculating Empirical Formula
  • A 5.325-g sample of methyl benzoate, a compound used in the manufacture of perfumes, contains 3.758 g of carbon, 0.316 g of hydrogen, and 1.251 g of oxygen. What is the empirical formula of this substance?
  • Answer: C4H4O
  • Practice Exercise
  • Solution (continued)
  • The ratio for H is too far from 1 to attribute the difference to experimental error; in fact, it is quite close to 1 1/3. This suggests that if we multiply the ratio by 3, we will obtain whole numbers:
  • The whole-number mole ratio gives us the subscripts for the empirical formula:
  • Check It is reassuring that the subscripts are moderately sized whole numbers. Otherwise, we have little by which to judge the reasonableness of our answer.
  • Sample Exercise 2.15 Determining Empirical Formula by Combustion Analysis
  • To find the empirical formula, we must compare the relative number of moles of each element in the sample. The relative number of moles of each element is found by dividing each number by the smallest number, 0.0043. The mole ratio of C:H:O so obtained is 2.98:7.91:1.00. The first two numbers are very close to the whole numbers 3 and 8, giving the empirical formula C3H8O.
  • Check The subscripts work out to be moderately sized whole numbers, as expected.
  • Solution (continued)
  • The calculation of the number of grams
  • of H from the grams of H2O is similar, although
  • we must remember that there are
  • 2 mol of H atoms per 1 mol of H2O
  • molecules:
  • The total mass of the sample, 0.255 g, is the
  • sum of the masses of the C, H, and O. Thus,
  • we can calculate the mass of O as follows:
  • We then calculate the number of moles of C,
  • H, and O in the sample:
  • Sample Exercise 2.16 Calculating Amounts of Reactants and Products
  • Balance this equation: C6H12O6(s) + O2(g)→ CO2(g) + H2O(l)
  • What is the type of this reaction?
  • How many grams of water are produced in the oxidation of 1.00 g of glucose, C6H12O6
  • Solution
  • C6H12O6(s) + 6O2(g)→ 6CO2(g) + 6 H2O(l)
  • This is a combustion reaction
  • 1 mol C6H12O6 6 mol H2O
  • Determine the molar mass of C6H12O6 and H2O, M C6H12O6 = 180.0 g/mol, M H2O = 18.0 g/mol
  • 180.0 g C6H12O6 6 X18.0 g = 108.0 g H2O
  • 1.00 g C6H12O6 1.00 x 108.0 = 0.600 g H2O
  • 180.0
  • Answer : The oxidation of 1.00 g of glucose, C6H12O6 produces 0.600 g of water, H2O
  • 3) Analyze We are given the mass of a reactant and are asked to determine the mass of a product in the given equation.
  • From the balanced equation :C6H12O6(s) + 6O2(g)→ 6CO2(g) + 6 H2O(l)
  • Sample Exercise 2.16 Calculating Amounts of Reactants and Products
  • The decomposition of KClO3 is commonly used to prepare small amounts of O2 in the laboratory:
  • 2 KClO3(s) → 2 KCl(s) + 3 O2(g). How many grams of O2 can be prepared from 4.50 g of KClO3?
  • Answer: 1.77 g
  • Practice Exercise
  • Solution (continued)
  • Check An estimate of the magnitude of our answer, 18/180 = 0.1 and 0.1  6 = 0.6, agrees with the exact calculation. The units, grams H2O, are correct. The initial data had three significant figures, so three significant figures for the answer is correct.
  • Comment An average person ingests 2 L of water daily and eliminates 2.4 L. The difference between 2 L and 2.4 L is produced in the metabolism of foodstuffs, such as in the oxidation of glucose. (Metabolism is a general term used to describe all the chemical processes of a living animal or plant.) The desert rat (kangaroo rat), on the other hand, apparently never drinks water. It survives on its metabolic water.
  • Sample Exercise 2.17 Calculating Amounts of Reactants and Products
  • Solid lithium hydroxide is used in space vehicles to remove the carbon dioxide exhaled by astronauts. The lithium hydroxide reacts with gaseous carbon dioxide to form solid lithium carbonate and liquid water. How many grams of carbon dioxide can be absorbed by 1.00 g of lithium hydroxide?
  • Solution
  • Analyze We are given a verbal description of a reaction and asked to calculate the number of grams of one reactant that reacts with 1.00 g of another.
  • Plan The verbal description of the reaction can be used to write a balanced equation:
  • 2 LiOH(s) + CO2(g) → Li2CO3(s) + H2O(l)
  • We are given the grams of LiOH and asked to calculate grams of CO2. We can accomplish this task by using the following sequence of conversions:
  • Grams LiOH → moles LiOH → moles CO2 → grams CO
  • The conversion from grams of LiOH to moles of LiOH requires the molar mass of LiOH (6.94 + 16.00 + 1.01 = 23.95 g/mol). The conversion of moles of LiOH to LiOH (6.94 + 16.00 + 1.01 = 23.95 g/mol) moles of CO2 is based on the balanced chemical equation: 2 mol LiOH 1 mol CO2. To convert the number of moles of CO2 to grams, we must use the molar mass of CO2: 12.01 + 2(16.00) = 44.01 g/mol.
  • Sample Exercise 2.17 Calculating Amounts of Reactants and Products
  • Propane, C3H8, is a common fuel used for cooking and home heating. What mass of O2 is consumed in the combustion of 1.00 g of propane?
  • Answer: 3.64 g
  • Practice Exercise
  • Solution (continued)
  • Check Notice that 23.95 ≈ 24, 24  2 = 48, and 44/48 is slightly less than 1. The magnitude of the answer is reasonable based on the amount of starting LiOH; the significant figures and units are appropriate, too.
  • The most important commercial process for converting N2 from the air into nitrogen-containing compounds is based on the reaction of N2 and H2 to form ammonia (NH3):
  • N2(g) + 3 H2(g)→2 NH3(g)
  • How many moles of NH3 can be formed from 3.0 mol of N2 and 6.0 mol of H2?
  • Solution
  • Analyze We are asked to calculate the number of moles of product, NH3, given the quantities of each reactant, N2 and H2, available in a reaction. Thus, this is a limiting reactant problem.
  • Plan If we assume that one reactant is completely consumed, we can calculate how much of the second reactant is needed in the reaction. By comparing the calculated quantity with the available amount, we can determine which reactant is limiting. We then proceed with the calculation, using the quantity of the limiting reactant.
  • Solve The number of moles of H2 needed for complete consumption of 3.0 mol of N2 is:
  • Because only 6.0 mol H2 is available, we will run out of
  • H2 before the N2 is gone, and H2 will be the limiting reactant. We use the quantity of the limiting reactant, H2, to calculate the quantity of NH3 produced:
  • Comment The table on the right summarizes
  • this example:
  • Sample Exercise 2.18 Calculating the Amount of Product Formed from
  • a Limiting Reactant
  • Consider the reaction 2 Al(s) + 3 Cl2(g) → 2 AlCl3(s). A mixture of 1.50 mol of Al and 3.00 mol of Cl2 is allowed to react. (a) Which is the limiting reactant? (b) How many moles of AlCl3 are formed? (c) How many moles of the excess reactant remain at the end of the reaction?
  • Answers: (a) Al, (b) 1.50 mol, (c) 0.75 mol Cl2
  • Practice Exercise
  • Solution (continued)
  • Notice that we can calculate not only the number of moles of NH3 formed but also the number of moles of each of the reactants remaining after the reaction. Notice also that although the number of moles of H2 present at the beginning of the reaction is greater than the number of moles of N2 present, the H2 is nevertheless the limiting reactant because of its larger coefficient in the balanced equation.
  • Check The summarizing table shows that the mole ratio of reactants used and product formed conforms to the coefficients in the balanced equation, 1:3:2. Also, because H2 is the limiting reactant, it is completely consumed in the reaction, leaving 0 mol at the end. Because 6.0 mol H2 has two significant figures, our answer has two significant figures.
  • Sample Exercise 2.19 Calculating the Amount of Product Formed from
  • a Limiting Reactant
  • Consider the following reaction that occurs in a fuel cell:
  • 2 H2(g) + O2 (g) → 2 H2O (g)
  • This reaction, properly done, produces energy in the form of electricity and water. Suppose a fuel cell is set up with 150 g of hydrogen gas and 1500 grams of oxygen gas (each measurement is given with two significant figures). How many grams of water can be formed?
  • Solution
  • Analyze We are asked to calculate the amount of a product, given the amounts of two reactants, so this is a limiting reactant problem.
  • Plan We must first identify the limiting reagent. To do so, we can calculate the number of moles of each reactant and compare their ratio with that required by the balanced equation. We then use the quantity of the limiting reagent to calculate the mass of water that forms.
  • Solve From the balanced equation, we have the following stoichiometric relations:
  • Using the molar mass of each substance, we can calculate the number of moles of each reactant:
  • Sample Exercise 2.19 Calculating the Amount of Product Formed from
  • a Limiting Reactant
  • A strip of zinc metal with a mass of 2.00 g is placed in an aqueous solution containing 2.50 g of silver nitrate, causing the following reaction to occur:
  • Zn(s) + 2 AgNO3(aq) → 2 Ag(s) + Zn(NO3)2(aq)
  • Practice Exercise
  • Solution (continued)
  • Thus, there are more moles of H2 than O2. The coefficients in the balanced equation indicate, however, that the reaction requires 2 moles of H2 for every 1 mole of O2. Therefore, to completely react all the O2, we would need 2  47 = 94 moles of H2. Since there are only 75 moles of H2, H2 is the limiting reagent. We therefore use the quantity of H2 to calculate the quantity of product formed. We can begin this calculation
  • with the grams of H2, but we can save a step by starting with the moles of H2 that were calculated previously in the exercise:
  • Check The magnitude of the answer seems reasonable. The units are correct, and the number of significant figures (two) corresponds to those in the numbers of grams of the starting materials.
  • Comment The quantity of the limiting reagent, H2, can also be used to determine the quantity of O2 used (37.5 mol = 1200 g). The number of grams of the excess oxygen remaining at the end of the reaction equals the starting amount minus the amount consumed in the reaction, 1500 g – 1200 g = 300 g.
  • Sample Exercise 2.19 Calculating the Amount of Product Formed from
  • a Limiting Reactant
  • (a) Which reactant is limiting? (b) How many grams of Ag will form? (c) How many grams of Zn(NO3)2 will form? (d) How many grams of the excess reactant will be left at the end of the reaction?
  • Answers: (a) AgNO3, (b) 1.59 g, (c) 1.39 g, (d) 1.52 g Zn
  • Practice Exercise (continued)
  • Sample Exercise 2.20 Calculating the Theoretical Yield and the Percent
  • Yield for a Reaction
  • Adipic acid, H2C6H8O4, is used to produce nylon. The acid is made commercially by a controlled reaction between cyclohexane (C6H12) and O2:
  • 2 C6H12(l) + 5 O2(g) → 2 H2C6H8O4(l) + 2 H2O(g)
  • (a) Assume that you carry out this reaction starting with 25.0 g of cyclohexane and that cyclohexane is the limiting reactant. What is the theoretical yield of adipic acid?
  • (b) If you obtain 33.5 g of adipic acid from your reaction, what is the percent yield of adipic acid?
  • Solution
  • Analyze We are given a chemical equation and the quantity of the limiting reactant (25.0 g of C6H12). We are asked first to calculate the theoretical yield of a product (H2C6H8O4) and then to calculate its percent yield if only 33.5 g of the substance is actually obtained.
  • Plan (a) The theoretical yield, which is the calculated quantity of adipic acid formed in the reaction, can be calculated using the following sequence of conversions:
  • g C6H12 → mol C6H12 → mol H2C6H8O4 → g H2C6H8O4
  • Sample Exercise 2.20 Calculating the Theoretical Yield and the Percent
  • Yield for a Reaction
  • Solution (continued)
  • (b) The percent yield is calculated by comparing the actual yield (33.5 g) to the theoretical yield using Equation 3.14.
  • Solve
  • Check Our answer in (a) has the appropriate magnitude, units, and significant figures. In (b) the answer is less than 100% as necessary.
  • Imagine that you are working on ways to improve the process by which iron ore containing Fe2O3 is converted into iron. In your tests you carry out the following reaction on a small scale:
  • Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)
  • (a) If you start with 150 g of Fe2O3 as the limiting reagent, what is the theoretical yield of Fe? (b) If the actual yield of Fe in your test was 87.9 g, what was the percent yield?
  • Answers: (a) 105 g Fe, (b) 83.7%
  • Practice Exercise
  • Sample Exercise 2.21 Calculating Molarity
  • Calculate the molarity of a solution made by dissolving 23.4 g of sodium sulfate (Na2SO4) in enough water to form 125 mL of solution.
  • Check: Because the numerator is only slightly larger than the denominator, it is reasonable for the answer to be a little over 1 M. The units (mol/L) are appropriate for molarity, and three significant figures are appropriate for the answer because each of the initial pieces of data had three significant figures.
  • Solution
  • Analyze: We are given the number of grams of solute (23.4 g), its chemical formula (Na2SO4), and the volume of the solution (125 ml). We are asked to calculate the molarity of the solution.
  • Plan: We can calculate molarity using Equation 4.33. To do so, we must convert the number of grams of solute to moles and the volume of the solution from milliliters to liters.
  • Solve: The number of moles of Na2SO4 is obtained by using its molar mass:
  • Converting the volume of the solution to liters:
  • Thus, the molarity is
  • Sample Exercise 2.21 Calculating Molarity
  • Calculate the molarity of a solution made by dissolving 5.00 g of glucose (C6H12O6) in sufficient water to form exactly 100 mL of solution.
  • Answer: 0.278 M
  • Practice Exercise
  • Sample Exercise 3.22 Calculating Molar Concentrations of Ions
  • What are the molar concentrations of each of the ions present in a 0.025 M aqueous solution of
  • calcium nitrate?
  • What is the molar concentration of K+ ions in a 0.015 M solution of potassium carbonate?
  • Answer: 0.030 M K+
  • Practice Exercise
  • Solution
  • Analyze: We are given the concentration of the ionic compound used to make the solution and asked to determine the concentrations of the ions in the solution.
  • Plan: We can use the subscripts in the chemical formula of the compound to determine the relative concentrations of the ions.
  • Solve: Calcium nitrate is composed of calcium (Ca2+) ions and nitrate ions NO3–, so its chemical formula is Ca(NO3)2. Because there are two NO3– ions for each Ca2+ ion in the compound, each mole of Ca(NO3)2 that dissolves dissociates into 1 mol of Ca2+ and 2 mol of NO3–. Thus, a solution that is 0.025 M in Ca(NO3)2 is 0.025 M in Ca2+ and 2  0.025 M = 0.050 M in NO–:
  • Check: The concentration of NO3– ions is twice that of Ca2+ ions, as the subscript 2
  • after the NO3– in the chemical formula Ca(NO3)2 suggests it should be.
  • Sample Exercise 3.23 Using Molarity to Calculate Grams of Solute
  • How many grams of Na2SO4 are required to make 0.350 L of 0.500 M Na2SO4?
  • Solution
  • Analyze: We are given the volume of the solution (0.350 L), its concentration (0.500 M), and the identity of the solute Na2SO4 and asked to calculate the number of grams of the solute in the solution.
  • Plan: We can use the definition of molarity (Equation 4.33) to determine the number of moles of solute, and then convert moles to grams using the molar mass of the solute.
  • Solve: Calculating the moles of Na2SO4 using the molarity and volume of solution gives
  • Because each mole of Na2SO4 weighs 142 g, the required number of grams of Na2SO4 is
  • Check: The magnitude of the answer, the units, and the number of significant figures are all appropriate.
  • Sample Exercise 3.23 Using Molarity to Calculate Grams of Solute
  • (a) How many grams of Na2SO4 are there in 15 mL of 0.50 M Na2SO4? (b) How many milliliters of 0.50 M Na2SO4 solution are needed to provide 0.038 mol of this salt?
  • Answers: (a) 1.1 g, (b) 76 mL
  • Practice Exercise
  • Sample Exercise 3.24 Calculation of Mass-Related Concentrations
  • (a) Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water. (b) A commercial bleaching solution contains 3.62 mass % sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2.50 kg of bleaching solution?
  • Answer: (a) 2.91%, (b) 90.5 g of NaOCl
  • Practice Exercise
  • (a) A solution is made by dissolving 13.5 g of glucose (C6H12O6) in 0.100 kg of water. What is the mass percentage of solute in this solution? (b) A 2.5-g sample of groundwater was found to contain 5.4 µg of Zn2+. What is the concentration of Zn2+ in parts per million?
  • Solution
  • (a) Analyze: We are given the number of grams of solute (13.5 g) and the number of grams of solvent (0.100 kg = 100 g). From this we must calculate the mass percentage of solute.
  • Plan: We can calculate the mass percentage by using Equation 13.5. The mass of the solution is the sum of the mass of solute (glucose) and the mass of solvent (water).
  • Comment: The mass percentage of water in this solution is (100 – 11.9)% = 88.1%.
  • (b) Analyze: In this case we are given the number of micrograms of solute. Because 1 µg is 1 × 10–6 g,
  • 5.4 µg = 5.4 × 10–6 g.
  • Plan: We calculate the parts per million using Equation 13.6.
  • Sample Exercise 3.25 Calculation of Molality
  • A solution is made by dissolving 4.35 g glucose (C6H12O6) in 25.0 mL of water at 25 °C. Calculate the molality of glucose in the solution. Water has a density of 1.00 g/mL.
  • What is the molality of a solution made by dissolving 36.5 g of naphthalene (C10H8) in 425 g of toluene (C7H8)?
  • Answer: 0.670 m
  • Practice Exercise
  • Solution
  • Analyze: We are asked to calculate a molality. To do this, we must determine the number of moles of solute (glucose) and the number of kilograms of solvent (water).
  • Plan: We use the molar mass of C6H12O6 to convert grams to moles. We use the density of water to convert milliliters to kilograms. The molality equals the number of moles of solute divided by the number of kilograms of solvent (Equation 13.9).
  • Solve: Use the molar mass of glucose, 180.2 g/mL, to convert grams to moles:
  • Because water has a density of 1.00 g/mL, the mass of the solvent is
  • Finally, use Equation 13.9 to obtain the molality:
  • Sample Exercise 3.26 Calculation of Mole Fraction and Molality
  • An aqueous solution of hydrochloric acid contains 36% HCl by mass. (a) Calculate the mole fraction of HCl in the solution. (b) Calculate the molality of HCl in the solution.
  • Solution
  • Analyze: We are asked to calculate the concentration of the solute, HCl, in two related concentration units, given only the percentage by mass of the solute in the solution.
  • Plan: In converting concentration units based on the mass or moles of solute and solvent (mass percentage, mole fraction, and molality), it is useful to assume a certain total mass of solution. Let’s assume that there is exactly 100 g of solution. Because the solution is 36% HCl, it contains 36 g of HCl and (100 – 36) g = 64 g of H2O. We must convert grams of solute (HCl) to moles to calculate either mole fraction or molality. We must convert grams of solvent (H2O) to moles to calculate mole fractions, and to kilograms to calculate molality.
  • Solve: (a) To calculate the mole fraction of HCl, we convert the masses of HCl and H2O to moles and then use Equation 13.7:
  • (b) To calculate the molality of HCl in the solution,
  • we use Equation 13.9. We calculated the number of moles of HCl in part (a), and the mass of solvent is 64 g = 0.064 kg:
  • Sample Exercise 3.26 Calculation of Mole Fraction and Molality
  • Acommercial bleach solution contains 3.62 mass % NaOCl in water. Calculate (a) the mole fraction and (b) the molality of NaOCl in the solution.
  • Answer: (a) 9.00 × 10–3, (b) 0.505 m.
  • Practice Exercise
  • Sample Exercise 3.27 Calculation of Molarity Using the Density of a Solution
  • A solution with a density of 0.876 g/mL contains 5.0 g of toluene (C7H8) and 225 g of benzene. Calculate the molarity of the solution.
  • The units for our answer (mol/L) are correct, and the answer, 0.21 M, has two significant figures, corresponding to the number of significant figures in the mass of solute (2).
  • Solution
  • Analyze: Our goal is to calculate the molarity of a solution, given the masses of solute (5.0 g) and solvent (225 g) and the density of the solution (0.876 g/mL).
  • Plan: The molarity of a solution is the number of moles of solute divided by the number of liters of solution (Equation 13.8). The number of moles of solute (C7H8) is calculated from the number of grams of solute and its molar mass. The volume of the solution is obtained from the mass of the solution (mass of solute + mass of solvent = 5.0 g + 225 g = 230 g) and its density.
  • Solve: The number of moles of solute is
  • The density of the solution is used to convert the mass of the solution to its volume:
  • Molarity is moles of solute per liter of solution:
  • Check: The magnitude of our answer is reasonable. Rounding moles to 0.05 and liters to 0.25 gives a molarity of
  • Sample Exercise 3.27 Calculation of Molality Using the Density of a Solution
  • A solution containing equal masses of glycerol (C3H8O3) and water has a density of 1.10 g/mL. Calculate (a) the molality of glycerol, (b) the mole fraction of glycerol, (c) the molarity of glycerol in the solution.
  • Answer: (a) 10.9 m, (b) = 0.163, (c) 5.97 M
  • Practice Exercise
  • Solution (continued)
  • Comment: Because the mass of the solvent
  • (0.225 kg) and the volume of the solution (0.263 L)
  • are similar in magnitude, the molarity and molality
  • are also similar in magnitude:
  • Sample Exercise 3.28 Preparing A solution by Dilution
  • How many milliliters of 3.0 M H2SO4 are needed to make 450 mL of 0.10 M H2SO4?
  • Solution
  • Analyze: We need to dilute a concentrated solution. We are given the molarity of a more concentrated solution (3.0 M) and the volume and molarity of a more dilute one containing the same solute (450 mL of 0.10 M solution). We must calculate the volume of the concentrated solution needed to prepare the dilute solution.
  • Plan: We can calculate the number of moles of solute, H2SO4, in the dilute solution and then calculate the volume of the concentrated solution needed to supply this amount of solute. Alternatively, we can directly apply Equation 4.35. Let’s compare the two methods.
  • Solve: Calculating the moles of H2SO4 in the dilute solution:
  • Calculating the volume of the concentrated solution that contains 0.045 mol H2SO4:
  • Converting liters to milliliters gives 15 mL.
  • If we apply Equation 4.35, we get the same result:
  • Sample Exercise 3.28 Preparing A solution by Dilution
  • (a) What volume of 2.50 M lead(II) nitrate solution contains 0.0500 mol of Pb2+?
  • (b) How many milliliters of 5.0 M K2Cr2O7 solution must be diluted to prepare 250 mL of 0.10 M solution?
  • (c) If 10.0 mL of a 10.0 M stock solution of NaOH is diluted to 250 mL, what is the concentration of the resulting stock solution?
  • Answers: (a) 0.0200 L = 20.0 mL, (b) 5.0 mL, (c) 0.40 M
  • Practice Exercise
  • Solution (continued)
  • Either way, we see that if we start with 15 mL of 3.0 M H2SO4 and dilute it to a total volume of 450 mL, the desired 0.10 M solution will be obtained.
  • Check: The calculated volume seems reasonable because a small volume of concentrated solution is used to prepare a large volume of dilute solution.


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