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Balance the following representation of the chemical reaction involved when an airbag deploys.

EXAMPLE 6.1. Balancing Equations. Solution.

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Balance the following representation of the chemical reaction involved when an airbag deploys.

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  1. EXAMPLE 6.1 Balancing Equations Solution Sodium atoms are balanced, but the nitrogen atoms are not. For this sort of problem, we will use the concept of the least common multiple. There are three nitrogen atoms on the left (reactants side) and two on the right (products side). The least common multiple of 2 and 3 is 6. Therefore, we need three N2 and two NaN3: 2 NaN3 Na + 3 N2 (not balanced) We now have two sodium atoms on the left. We can get two on the right by placing the coefficient 2 in front of Na. 2 NaN3 2 Na + 3 N2 (balanced) Checking, we count two Na atoms and six N atoms on each side. The equation is balanced. 2 x 3 = 6 N atoms 3 x 2 = 6 N atoms 2 Na atoms 1 Na atoms Balance the following representation of the chemical reaction involved when an airbag deploys. NaN3 Na + N2

  2. EXAMPLE 6.1 Balancing Equations continued Exercise 6.1A Exercise 6.1B The reaction between hydrogen and nitrogen to give ammonia, called the Haber process, is typically the first step in the industrial production of nitrogen fertilizers, represented as H2 + N2 NH3 Balance the equation. Iron ores like Fe2O3 are smelted by reaction with carbon to produce metallic iron and carbon dioxide, represented as Fe2O3 + C  CO3 + Fe Balance the equation.

  3. EXAMPLE 6.2 Balancing Equations Solution In this equation, oxygen appears in two different products and by itself in 02; we leave the oxygen for last and balance the other two elements first. Carbon is already balanced, with one atom on each side of the equation. For hydrogen, the least common multiple of 2 and 4 is 4, and so we place the coefficient 2 in front of H2O to balance hydrogen. Now we have four hydrogen atoms on each side. CH4 + O2 CO2+ 2 H2O (not balanced) Now for the oxygen. There are four oxygen atoms on the right. If we place a 2 in front of O2 on the left, the oxygen atoms balance. CH4 + 2 O2 CO2+ 2 H2O (balanced) The equation now has one C atom, four H atoms, and four O atoms on each side; it is balanced. When a fuel such as methane is burned in sufficient air, the products are carbon dioxide and water, represented as CH4 + O2  CO2 + H2O (not balanced) Balance the equation.

  4. EXAMPLE 6.2 Balancing Equations continued Exercise 6.2A Exercise 6.2B Butane is a common fuel. Its combustion is represented as C4H10 + O2 CO2+ H2O Balance the equation. Does it take more oxygen (per molecule) to burn butane (see Exercise 6.2A) than it does to burn methane? How much more CO2 is produced?

  5. EXAMPLE 6.3 Volume Relationships of Gases Solution The coefficients in the equation indicate that each volume of C3H8(g) requires 5 volumes of O2(g). Thus, we use 5 L O2(g)/1 L C3H8(g) as the ratio to find the volume of oxygen required. ? L O2(g) = O.556 L C3 H8(g) x = 2.78 L O2(g) 5 L O2(g) Exercise 6.3A Exercise 6.3B 1 L C3 H8(g) Using the equation in Example 6.3, calculate the volume of CO2(g) produced when 0.492 L of propane is burned if the two gases are compared at the same temperature and pressure. If 10.0 L each of propane and oxygen are combined at the same temperature and pressure, which gas will be left over after reaction? What volume of that gas will remain? What volume of oxygen is required to burn 0.556 L of propane if both gases are measured at the same temperature and pressure? C3H8(g)+ 5 O2(g)  3 CO2 (g) + 4 H2O(g)

  6. EXAMPLE 6.4 Calculating Molecular Masses Solution We start with the molecular formula: NO2. Then, to determine the molecular mass, we need only to add the atomic mass of nitrogen to twice the atomic mass of oxygen. Using a calculator, we need only write down the final answer, 46.0 u. That is, we have no need to record the numbers 14.0 and 32.0. We must make certain that all the atoms in the formula unit are accounted for, which means paying particular attention to all the subscripts and parentheses in the formula. The “(NH4)2” means that both the “N” and the “H4” must be multiplied by 2—that is, the formula indicates a total of two N atoms and eight H atoms. Combining the atomic masses, we have 2 x atomic mass of N = 2 x 14.0 u = 28.0 u 8 x atomic mass of H = 8 x 1.01 u = 8.08 u 1 x atomic mass of S = 1 x 32.0 u = 32.0 u 4 x atomic mass of O = 4 x 16.0 u = 64.0 u Formula mass of (NH4)2SO4 = 132.1 u 1 x atomic mass of N = 1 x 14.0 u = 14.0 u 2 x atomic mass of O = 2 x 16.0 u = 32.0 u Formula mass of NO2 = 46.0 u Calculate (a) the molecular mass of nitrogen dioxide (NO2) an amber colored gas that is a constituent of smog, and (b) the formula mass of ammonium sulfate [(NH4)2SO4] a fertilizer commonly used by home gardeners.

  7. EXAMPLE 6.4 Calculating Molecular Masses continued Exercise 6.4A Exercise 6.4B Calculate the formula mass of (a) sodium azide (NaN3) used in automobile airbags, and (b) phosphoric acid (H3PO4). Calculate the formula mass of (a)para-dichlorobenzene (C6H4Cl2) used as a moth repellent, and (b) calcium dihydrogen phosphate [Ca(H2PO4)2], used as a mineral supplement in foods.

  8. EXAMPLE 6.5 Mole-to-Mass Conversions Solution 28.0 g N2 The molecular mass of N2 is 2 x 14.0 u = 28.0 u. The molar mass of N2 is therefore 28.0 g/mol. Using the molar mass as a conversion factor (red), we have ? g N2 = 0.400 mol N2x = 11.2 g N2 1 mol N2 Exercise 6.5 Calculate the mass, in grams, of (a) 0.0728 mol silicon, (b) 55.5 mol H2O, and (c) 0.0728 mol Ca(H2PO4)2. How many grams of N2 are in 0.400 moles N2?

  9. EXAMPLE 6.6 Mass-to-Mole Conversions Solution 1 mol Na The molar mass of Na is 23.0 g/mol. To convert from a mass in grams to an amount in moles, we must use the inverse of the molar mass as a conversion factor (1 mol Na/23.0 g Na) to get the proper cancellation of units. When we start with grams, we must have grams in the denominator of our conversion factor (red). ? mol Na = 62.5 g Na x = 2.72 mol Na 23.0 g Na Exercise 6.6 Calculate the amount, in moles, of (a) 3.71 g Fe, (b) 165 g butatne, C4H10, and (c) 0.100 mol Mg(NO3)2. Calculate the number of moles of Na in a 62.5-g sample of sodium metal.

  10. EXAMPLE 6.7 Density of a Gas at STP Solution 28.0 g N2 1 mol N2 The molar ass of N2 gas is 28.0 g/mol. We multiply by the conversion factor 1 mol N2 = 22.4 L, arranged to cancel units of moles. b. The molar mass of CH4 gas is (1 x 12.0) g/mol + (4 x 1.01) g/mol = 16.0 g/mol. Again we use the conversion factor 1 mol CH4 = 22.4 L. x = 1.25 g/L 1 mol N2 22.4 L N2 Exercise 6.7A Exercise 6.7B Calculate the density of He at STP. Estimate the density of air at STP (assume 78% N2 and 22% O2) and compare this value to the value of He you calculated in Exercise 6.7A. 16.0 g CH4 1 mol CH4 x = 0.714 g/L 1 mol CH4 22.4 L CH4 Calculate the density of (a) nitrogen gas and (b) methane (CH4) gas, both at STP.

  11. EXAMPLE 6.8 Molar Mass from Gas Densities Solution This time we are given the density in g/L. We want molar mass, which has units of g/mol. Once again we have the factor 22.4 L = 1 mol diethyl ether. Clearly, we must cancel the unit of liters and obtain the unit of moles in the denominator. 3.30 g 22.4 L x = 73.9 g/mol 1 L 1 mol Exercise 6.8A Exercise 6.8B The density of an unknown gas at STP is 2.30 g/L. Calculate its molar mass. An unknown gaseous compound contains only hydrogen and carbon and its density is 1.34 g/mol at STP. What is the formula for the compound? The density of diethyl ether vapor is 3.30 g/L. Calculate the molar mass of diethyl ether.

  12. Molecular, Molar, and Mass Relationships CONCEPTUAL EXAMPLE 6.9 Solution The molecular and molar relationships can be obtained directly from the equation; no calculation is necessary. The mass relationship requires a little calculation: Molecular: Two molecules of NO react with one molecule of O2 to form two molecules of NO2. Molar: 2 mol of NO reacts with 1 mol of O2 to form 2 mol of NO2. Mass: 60.0 g of NO (2 mol NO x 30.0 g/mol) reacts with 32.0 g (1 mol O2x 32.0 g/mol) of O2 to form 92.0 g (2 mol NO2x 46.0 g/mol) of NO2. Exercise 6.9 Hydrogen sulfide, a gas that smells like rotten eggs, burns in air to produce sulfur dioxide and water according to the equation 2 H2S + 3 O2 2 SO2 + H2O State the molecular, molar, and mass relationships indicated by this equation. Nitrogen monoxide (nitric oxide), an air pollutant discharged by internal combustion engines, combines with oxygen to form nitrogen dioxide, a yellowish-brown gas that irritates the respiratory system and eyes. The equation for this reaction is 2 NO + O2 2 NO2 State the molecular, molar, and mass relationships indicated by the equation.

  13. EXAMPLE 6.10 Molar Relationships Solution The equation tells us that 5 mol O2 is required to burn 1 mol C3H8. We can write 1 mol C3H8 5 mol O2 where we use the symbol to mean “is stoichiometrically equivalent to.” From this relationship we can construct conversion factors to relate moles of oxygen to moles of propane. The possible conversion factors are Which one do we use? Only if we multiply the given quantity (0.105 mol C3H8) by the factor on the right do we get an answer with the asked-for units (moles of oxygen). ? mol O2 = 0.105 mol C3H8x = 0.525 mol O2 Exercise 6.10 For the combustion of propane in Example 6.10, (a) How many moles of carbon dioxide is formed when 0.529 mol of C3H8 is burned? (b) How many moles of water is produced when 76.2 mol of C3H8 is burned? (c) How many moles of carbon dioxide is produced when 1.010 mol of O2 is consumed? 1 mol C3H8 5 mol O2 and 5 mol O2 1 mol C3H8 5 mol O2 1 mol C3H8 When 0.105 mol of propane is burned in a plentiful supply of oxygen, how many moles of oxygen is consumed? C3H8 + 5 O2 3 CO2 + 4 H2O

  14. EXAMPLE 6.11 Mass Relationships Solution Step 1 The balanced equation is C + O2 CO2 Step 2 The molar masses are 2 x 16.0 = 32.0 g/mol for O2 and 12.0 g/mol for C. Step 3 We convert the mass of the given substance, carbon, to an amount in moles. ? mol C = 10.0 g C x = 0.833 mol C Step 4 We use coefficients from the balanced equation equation to establish the stoichiometric factor (red) that relates the amount of oxygen to that of carbon. 0.833 mol C x = 0.833 mol O2 Step 5 We convert from moles of oxygen to grams of oxygen. 0.833 mol O2x = 26.7 g O2 1 mol O2 1 mol O2 32.0 g O2 12.0 g C 1 mol C 1 mol O2 Calculate the mass of oxygen needed to react with 10.0 g of carbon in the reaction that forms carbon dioxide.

  15. EXAMPLE 6.11 Mass Relationships continued We can also combine the five steps into a single setup. Note that the units in the denominators of the conversion factors are chosen so that each cancels the unit in the numerator of the preceding term. (The slightly different answers are due to rounding in the intermediate steps.) This converts g C to mol C The answer: the number and the unit This releasemol C to mol O2 This convertsmol O2 to g O2 We start here 1 mol O2 10.0 g C x x x = 26.6 g O2 1 mol C Exercise 6.11B Exercise 6.11A Calculate the mass of carbon dioxide formed by burning 775 g of each of (a) methane (CH4) and (b)butane (C4H10). Calculate the mass of oxygen (O2) needed to react with 0.334 g of nitrogen (N2) in the reaction that forms nitrogen dioxide. 1 mol C 32.0 g O2 1 mol O2 12.0 g C

  16. EXAMPLE 6.12 Mass Relationships Solution We start by writing and balancing the chemical equation, which shows that 2 mol NaN3 produces 2 mol Na and 3 mol N2. 2 NaN3 2 Na + 3 N2 The molar mass of NaN3 is 23.0 g/mol + (3 x 14.0) g/mol = 65.0 g/mol and the molar mass of N2 is (2 x 14.0) g/mol = 28.0 g/mol. We convert the mass of the given substance, sodium azide, to an amount in moles. 60.0 g NaN3x = 0.923 mol NaN3 We use coefficients from the balanced equation to establish the stoichiometric factor that relates the amount of nitrogen gas to that of sodium azide. 0.923 mol NaN3x = 1.38 mol N2 1 mol NaN3 3 mol N2 65.0 g NaN3 2 mol NaN3 The decomposition of sodium azide (NaN3) produces sodium metal and nitrogen gas. The gas is used to inflate automobile airbags. What mass of nitrogen, in grams, can be made from 60.0 g of sodium azide?

  17. EXAMPLE 6.12 Mass Relationships continued We convert from moles of nitrogen gas to grams of nitrogen gas. 1.38 mol N2x = 38.6 g N2 As is usually the case, all the steps just outlined can be combined into a single setup. 60.0 g NaN3x x x = 38.8 g NaN3 Notice that the units of the numerator in one stoichiometric factor are the units in the denominator of the next stoichiometric factor. In this way, the correct cancellation of units occurs and the units of the final numerator are the units of your answer. 1 mol NaN3 3 mol NaN3 28.0 g N2 65.0 g NaN3 2 mol NaN3 1 mol N2 Exercise 6.12A Exercise 6.12B Ammonia reacts with phosphoric acid (H3PO4) to form ammonium phosphate [(NH4)3PO4]. What mass in grams of ammonia is needed to react completely with 74.8 g of phosphoric acid? The decomposition of potassium chlorate (KCIO3) produces potassium chloride (KCI) and O2 gas. What mass in grams of oxygen can be made from 2.47 g of potassium chlorate? Phosphorus reacts with oxygen to form tetraphosphorus decoxide. The equation is P4 + O2 P4O10 (not balanced) What mass in grams of tetraphosphorus decoxide can be made from 3.50 g of phosphorus? 28.0 g N2 1 mol N2

  18. EXAMPLE 6.13 Boyle's Law: Pressure-Volume Relationships Solution The pressure increase from 0.524 atm to 5.15 atm is almost tenfold. The volume should drop to about one-tenth of the initial value. We estimate a volume of 0.20 L. (The calculated value is 0.203 L.) Exercise 6.13 A gas is enclosed in a 10.2-L tank at 1208 mmHg. (The mmHg is a pressure unit; 760 mmHg = 1 atm.) Which of the following is a reasonable value for the pressure when the gas is transferred to a 30.0-L tank? 300 mmHg 400 mmHg 3,600 mmHg 12,000 mmHg A gas is enclosed in a cylinder fitted with a piston. The volume of the gas is 2.00 L at 0.524 atm. The piston is moved to increase the gas pressure to 5.15 atm. Which of the following is a reasonable value for the volume of the gas at the greater pressure? 0.20 L 0.40 L 1.00 L 16.0 L

  19. EXAMPLE 6.14 Boyle’s Law: Pressure-Volume Relationships Solution We find it helpful to first separate the initial from the final condition. Initial Final ChangeP1 = 1470 psi P2 = 14.7 psiV1 = 2.25 L V2 = ? Then use the equation V1P1 = V2P2 and solve for the desired volume or pressure. In this case, we solve for V2. V2 = V2 = = 225 L Because the final pressure (14.7 psi) is less than the initial pressure (1470 psi), we expect the final volume (225 L) to be larger than the original volume (2.25 L), and we see that it is. The pressure goes down, therefore, the volume goes up. V1P1 P2 2.25 L x 1470 psi 14.7 psi A cylinder of oxygen has a volume of 2.25 L. The pressure of the gas is 1470 pounds per square inch (psi) at 20 °C. What volume will the oxygen occupy at standard atmospheric pressure (14.7 psi) assuming no temperature change?

  20. Exercise 6.14A Exercise 6.14B A sample of air occupies 73.3 mL at 98.7 atm and 0 ºC. What volume will the air occupy at 4.02 atm and 0 ºC? A sample of helium occupies 535 mL at 988 mmHg and 25 °C. If the sample is transferred to a 1.05-L flask at 25 °C, what will be the gas pressure in the flask? EXAMPLE 6.14 Boyle’s Law: Pressure-Volume Relationships continued

  21. Charles’s Law: Temperature-Volume Relationships EXAMPLE 6.15 Solution First, and most important, convert all temperatures to the Kelvin scale: T(K) = t(°C) + 273 The initial temperature (T1) in each case is (27 + 273) = 300 K and the final temperatures are (a) (47 + 273) = 320 K and (b) (–23 + 273) = 250 K. We start by separating the initial from the final condition. Initial Final Change t1 = 27 °C t2 = 47 °C T1 = 300 K T2 = 320 K V1 = 2.00 L V2 = ? Solving the equation = V1 V2 T1 T2 A balloon indoors, where the temperature is 27 °C, has a volume of 2.00 L. What would its volume be (a) in a hot room where the temperature is 47 °C, and (b) outdoors, where the temperature is –23 ºC? (Assume no change in pressure in either case.)

  22. Charles’s Law: Temperature-Volume Relationships continued EXAMPLE 6.15 for V2, we have V2 = V2 = = 2.13 L As expected, because the temperature increases, the volume must also increase. We have the same initial conditions as in (a), but different final conditions. Initial Final Change t1 = 27 °C t2 = -23 °C T1 = 300 K T2 = 250 K V1 = 2.00 L V2 = ? Again using Charles’s law, we solve the equation for V2: V2 = V2 = = 1.67 L As we expected, the volume decreased because the temperature decreased. V1T2 T1 2.00 L x 320 K 300 K V1T2 T1 2.00 L x 250 K 300 K

  23. Charles’s Law: Temperature-Volume Relationships continued EXAMPLE 6.15 Exercise 6.15A Exercise 6.15B A sample of oxygen gas occupies a volume of 2.10 L at 25 °C. What volume will this sample occupy at 150 °C? (Assume no change in pressure.) b. A sample of hydrogen occupies 692 L at 602 °C. If the pressure is held constant, what volume will the gas occupy after being cooled to 23 °C? At what Celsius temperature will the initial volume of oxygen in Exercise 6.15A occupy 0.750 L? (Assume no change in pressure.)

  24. EXAMPLE 6.16 Ideal Gas Law Solution a. We start by solving the ideal gas equation for V. V = V = x x 244 K = 20.0 L b. Here we solve the ideal gas equation for P. P = P = x x 303 K = 0.829 atm nRT P Exercise 6.16A 1.00 mol 0.0821 L • atm Determine (a) the pressure exerted by 0.0330 mol of oxygen in an 18.0-L container at 313 K, and (b) the volume occupied by 0.200 mol of nitrogen gas at 298 K and 0.980 atm. 1.00 atm mol • K nRT V 0.500 mol 0.0821 L • atm 15.0 L mol • K Use the ideal gas law to calculate (a) the volume occupied by 1.00 mol of nitrogen gas at 244 K and 1.00 atm pressure, and (b) the pressure exerted by 0.500 mol of oxygen in a 15.0-L container at 303 K.

  25. EXAMPLE 6.16 Ideal Gas Law continued Exercise 6.16B Determine the volume of nitrogen gas produced from the decomposition of 130 g sodium azide (about the amount in a typical automobile airbag) at 25 °C and 1 atm.

  26. EXAMPLE 6.17 Solution Concentration: Molarity Solution moles of solute 3.50 mol NaCl Molarity (M) = = = 1.75 M NaCl We read 1.75 M NaCl as “1.75 molar NaCl.” liters of solution 2.00 L solution Exercise 6.17A Exercise 6.17B Calculate the molarity of a solution that has 0.0500 mol of NH3 in 5.7 L of solution. Calculate the molarity of a solution made by dissolving 0.750 mol of H3PO4 in enough water to produce 775 mL of solution. Calculate the molarity of a solution made by dissolving 3.50 mol of NaCl in enough water to produce 2.00 L of solution.

  27. EXAMPLE 6.18 Solution Concentration: Molarity Solution First, we must convert grams of KHCO3 to moles of KHCO3 333 g KHCO3x = 3.33 mol KHCO3 Now use this value as the numerator in the defining equation for molarity. The solution volume, 10.0 L, is the denominator. 1 mol KHCO3 100.1 g KHCO3 Exercise 6.18 Calculate the molarity of each of the following solutions. 18.0 mol of H2SO4 in 2.00 L of solution 3.00 mol of KI in 2.39 L of solution c. 0.206 mol of HF in 752 mL of solution (HF is used for etching glass.) 3.33 mol KHCO3 Molarity = = 0.333 M KHCO3 10.0 L solution What is the molarity of a solution in which 333 g of potassium hydrogen carbonate is dissolved in enough water to make 10.0 L of solution?

  28. EXAMPLE 6.19 Solution Preparation: Molarity Solution 0.15 mol NaCl First we use the molarity as a conversion factor to calculate moles of NaCl. Then we use the molar mass to calculate the grams of NaCl. 0.500 L solution x = 0.075 mol NaCl 1 L solution Exercise 6.19A Exercise 6.19B What mass in grams of potassium hydroxide is required to prepare 2.00 L of 6.00 M KOH? What mass in grams of potassium hydroxide is required to prepare 100.0 mL of 1.00 M KOH? 58.4 g NaCl 0.75 mol NaCl x = 44 g NaCl 1 mol NaCl How many grams of NaCl is required to prepare 0.500 L of typical over-the-counter saline solution (about 0.15 M NaCl)?

  29. EXAMPLE 6.20 Moles from Molarity and Volume moles of solute 0.425 mol HCl Solution Liters of HCl solution = = = = 0.0354 L molarity 12.0 M HCl We would need 0.0354 L (35.4 mL) of the solution to have 0.425 mol. Remember that molarity is moles per liter of solution, not per liter of solvent. 0.425 mol HCl 12.0 mol HCl/L Exercise 6.20A Exercise 6.20B What volume in milliliters of 15.0 M aqueous ammonia (NH3) solution do you need to get 0.445 mol of NH3? What mass in grams of HNO3 is in 500 mL of rain that has a concentration of 2.0 x 10–5 M HNO3? Concentrated hydrochloric acid has a concentration of 12.0 M HCl. How many milliliters of this solution would one need to get 0.425 mol of HCl?

  30. EXAMPLE 6.21 Percent by Volume 120 mL oil Solution Percent by volumes = x 100% = 3.0% 4000 mL solution Exercise 6.21A Exercise 6.21B What is the volume percent of ethanol in a solution that has 58.0 mL water in 625 mL of an ethanol–water solution? Assume that the volumes are additive, and determine the volume percent toluene (C6H5CH3) in a solution made by mixing 40.0 mL of toluene with 75.0 mL of benzene (C6H6). Two-stroke engines use a mixture of 120 mL of oil dissolved in enough gasoline to make of 4.0 liters of fuel. What is the percent by volume of oil in this mixture?

  31. EXAMPLE 6.22 Solution Preparation: Percent by Volume We begin by rearranging the equation for percent by volume to solve for volume of solute. Substituting, we have Take 39 mL of acetic acid and add enough water to make 775 mL of solution. Notice that we don’t simply add 775 mL of water, because the final volume of solution must be 775 mL. Percent by volume x volume of solution Volume of solute = Solution 100% 5.0% x 775 mL = = 39 mL Exercise 6.22A Exercise 6.22B 100% Describe how to prepare 450 mL of an aqueous solution that is 70.0% isopropyl alcohol by volume. Describe how you would prepare exactly 2.00 L of an aqueous solution that is 9.77% acetic acid by volume. Describe how to make 775 mL of vinegar (about a 5.0% by volume solution of acetic acid in water).

  32. EXAMPLE 6.23 Percent by Mass Solution Use these values in the above percent-by-mass equation: 25.5 g NaCl Percent by mass = x 100% = 5.66% NaCl (25.5 + 425) g solution Exercise 6.23A Exercise 6.23B Hydrogen peroxide solutions for home use are 3.0% by mass solutions of H2O2 in water. What is the percent by mass of a solution of 9.40 g of H2O2 dissolved in 335 g (335 mL) of water? Sodium hydroxide (NaOH, lye) is used to make soap and is very soluble in water. What is the percent by mass of a solution that contains 1.00 kg of NaOH dissolved in 950 mL of water? What is the percent by mass of a solution of 25.5 g of NaCl dissolved in 425 g (425 mL) of water?

  33. EXAMPLE 6.24 Solution Preparation: Percent by Mass Solution We begin by rearranging the equation for percent by mass to solve for mass of solute. Substituting, we have Take 20.9 g of NaNO3 and add enough water to make 430 g of solution. percent by mass x mass of solution Mass of solute = 100 % 4.85% x 430 g = = 20.9 g Exercise 6.24A Exercise 6.24B 100 % Describe how you would prepare 125 g of an aqueous solution that is 4.50% glucose by mass. Describe how you would prepare 1750 g of isotonic saline, a commonly used intravenous (IV) solution that is 0.89% sodium chloride by mass. Describe how to make 430 g of an aqueous solution that is 4.85% by mass NaNO3.

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