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Section 12.5: Reactions of Gases and Gas Stoichiometry Tutorial 1 Practice, page 599 1. Given: volume of hydrogen, VH = 54 L 2 Required: volume of oxygen, VO 2 Solution: Step 1. Write the balanced chemical equation for the reaction, listing the given values and required values. 2 H 2 (g) + O 2 (g) ! 2 H 2O(g) 54 L VO 2 Step 2. Convert the volume of the given substance, VH , to the volume of the required substance, 2 VO . 2 VO = VH ! 2 2 = 54 L ! 1 molO 2 2 molH 2 1 molO 2 2 molH 2 VO = 27 L 2 Statement: The volume of oxygen required is 27 L. 2. Given: volume of ammonia, VNH = 34.5 L 3 Required: volume of hydrogen gas, VH 2 Solution: Step 1. Write the balanced chemical equation for the reaction, listing the given values and required values. 3 H 2 (g) + N 2 (g) ! 2 NH 3 (g) VH 34.5 L 2 Step 2. Convert the volume of the given substance, VNH , to volume of the required substance, 3 VH . 2 VH = VNH ! 2 3 3 molH 2 2 molNH = 34.5 L ! 3 3 molH 2 2 molNH 3 VH = 51.8 L 2 Statement: The volume of hydrogen gas required is 51.8 L. Copyright © 2011 Nelson Education Ltd. Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions 12.5-1 3. Given: volume of nitrogen, VN = 15.0 L 2 Required: volume of oxygen, VO 2 Solution: Step 1. Write the balanced chemical equation for the reaction, listing the given values and required values below. 4 KNO3 (s) ! 2 K 2O(g) + 2 N 2 (g) + 5 O 2 (g) 15.0 L VO 2 Step 2. Convert the volume of the given substance, VN , to volume of the required substance, 2 VO . 2 5 molO VO = VN ! 2 2 2 molN 2 = 15.0 L ! 2 5 molO 2 2 molN 2 VO = 37.5 L 2 Statement: The volume of oxygen required is 37.5 L. Tutorial 2 Practice, page 602 1. Given: mass of oxygen gas, mO = 128 g 2 molar mass of oxygen gas, M O = 32.00 g/mol 2 The conditions are standard temperature and pressure, so T = 273 K and P = 101.325 kPa. Required: volume of nitrogen dioxide, VNO 2 Solution: Step 1. Write the balanced chemical equation, listing the given and required quantities. 2 NO(g) + O 2 (g) ! 2 NO 2 (g) 128 g VNO 2 Step 2. Convert the mass of the given substance, mO , into an amount, nO . 2 nO = 128 g ! 2 2 1 mol 32.00 g nO = 4.00 mol 2 Step 3. Determine the amount of nitrogen dioxide produced from the amount of oxygen gas. 2 molNO 2 nNO = 4.00 mol ! 2 1 molO 2 nNO = 8.00 mol 2 Copyright © 2011 Nelson Education Ltd. Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions 12.5-2 Step 4. Determine the volume of the required substance, VNO , using the ideal gas law equation, PV = nRT. Remember that R = 8.314 kPa i L i mol PVNO = nNO RT 2 –1 –1 2 i K . 2 VNO = 2 = nNO RT 2 P (8.00 mol ) !" 8.314 kPa i L i mol –1 ( i K –1 # 273 K $ ) 101.325 kPa VNO = 179 L 2 Statement: When 128 g of oxygen gas reacts with nitrogen monoxide, 179 L of nitrogen dioxide is produced (under the given conditions of temperature and pressure). 2. Given: volume of carbon dioxide, VCO = 2.00 L 2 temperature, T = 300.0 K pressure, P = 101.3 kPa Required: mass of sodium carbonate, mNa CO 2 3 Solution: Step 1. Write the balanced chemical equation, listing the given and required quantities. Na 2CO3 (s) + 2 HCl(aq) ! 2 NaCl(aq) + CO 2 (g) + H 2O(g) mNa CO 2 2.00 L 3 Step 2. Determine the amount of the given substance, nCO , produced in the reaction using the 2 ideal gas law equation, PV = nRT. Remember that R = 8.314 kPa i L i mol–1 i K–1. PVCO = nCO RT 2 2 nCO = 2 = PVCO RT 2 (101.3 kPa )(2.00 L ) (8.314 kPa i L i mol –1 )( i K –1 300.0 K ) nCO = 0.0812 mol 2 Step 3. Determine the amount of the sodium carbonate that is required to react with the amount of carbon dioxide, nNa CO . 2 nNa CO = 0.0812 mol ! 2 3 3 1 molNa CO 2 1 molCO 3 2 nNa CO = 0.0812 mol 2 3 Step 4. Determine the mass of sodium carbonate that is required from the amount of sodium carbonate. Copyright © 2011 Nelson Education Ltd. Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions 12.5-3 M Na CO = 105.99 g/mol 2 3 105.99 g 1 mol 105.99 g = 0.0812 mol ! 1 mol = 8.61 g mNa CO = nNa CO ! 2 3 mNa CO 2 3 2 3 Statement: 8.61 g of sodium carbonate is required to produce 2.00 L of carbon dioxide at these conditions. 3. Given: volume of ethane, VC H = 82.0 L 2 6 temperature, t = 123 °C pressure, P = 105 kPa molar mass of ethane, M? Required: volume of oxygen, VO 2 Solution: Step 1. Convert the temperature value to kelvins. T = t + 273 = 123 + 273 T = 396 K Step 2. Write the balanced chemical equation, listing the given and required quantities. 2 C2 H 6 (g) + 7 O 2 (g) ! 6 H 2O (g) + 4 CO 2 (g) 82.0 L VO 2 Step 3. Convert the volume of the given substance, VC H , into an amount, nC H , using the ideal 2 6 gas law equation, PV = nRT. Remember that R = 8.314 kPa i L i mol PVC H = nC H RT 2 6 2 nC H = 2 6 = 2 –1 6 i K . 6 PVC H 2 6 RT (105 kPa )(82.0 L ) (8.314 kPa i L i mol nC H = 2.6152 mol 2 –1 6 –1 )( i K –1 396 K ) [two extra digits carried] Step 4. Determine the amount of oxygen required for the amount of ethane combusted. 7 molO 2 nO = 2.6152 mol ! 2 2 molC H 2 nO = 9.1532 mol 2 6 [two extra digits carried] Copyright © 2011 Nelson Education Ltd. Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions 12.5-4 Step 5. Determine the volume of the required substance, VO , using the ideal gas law equation, PV = nRT. Remember that R = 8.314 kPa i L i mol PVO = nO RT 2 –1 2 –1 i K . 2 VO = nNO RT 2 2 = P (9.1532 mol ) !" 8.314 kPa i L i mol –1 ( i K –1 # 396 K $ ) 105 kPa VO = 287 L 2 Statement: 287 L of oxygen is required for the combustion of 82.0 L of ethane at these conditions. Mini Investigation: Capturing Carbon Dioxide on a Small Scale, page 602 A. Answers may vary. Sample answer: Ca OH 2 (aq) + CO 2 (g) ! CaCO3 (s) + H 2O ( ) B. Answers may vary. Sample answer: The formation of a white precipitate, which is the calcium carbonate, supports the proposed chemical equation. C. Bromothymol blue is blue in the basic solution formed by calcium oxide. Carbon dioxide is an acidic oxide, so bubbling carbon dioxide into the mixture would form an acidic solution that changes the colour of bromothymol blue from blue to yellow. D. Answers may vary. Sample answer: Yes, this would be an effective method of removing carbon dioxide from the atmosphere, as long as the amount of carbon dioxide released into the atmosphere in the production of limewater is less than the amount captured by the process. E. Answers may vary. Sample answer: An experiment could be conducted in which ambient air is forced through the limewater and then the resulting solution is poured through a filter and the calcium carbonate precipitate is collected. The steps could then be repeated except with forcing exhaled air through the limewater. The two samples of precipitate collected could then be massed and compared to determine how many times greater the concentration of carbon dioxide in exhaled air is compared to ambient air. Copyright © 2011 Nelson Education Ltd. Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions 12.5-5 Section 12.5 Questions, page 603 1. (a) 2 C4 H10 (g) + 13 O 2 (g) ! 8 CO 2 + 10 H 2O (b) Given: volume of butane, VC H = 5.60 L 4 10 The conditions are standard temperature and pressure, so T = 273 K and P = 101.325 kPa. Required: volume of carbon dioxide, VCO 2 Solution: Step 1. Write the balanced chemical equation, listing the given and required quantities. 2 C4 H10 (g) + 13 O 2 (g) ! 8 CO 2 + 10 H 2O 5.60 L VCO 2 Step 2. Convert the volume of the given substance, VC H , into an amount, nC H . 4 10 4 10 1 mol 22.4 L = 0.250 mol nC H = 5.60 L ! 4 10 nC H 4 10 Step 3. Determine the amount of carbon dioxide produced from the amount of butane gas used. 8 molCO 2 nCO = 0.250 mol ! 2 2 molC H 4 10 nCO = 1.00 mol 2 Step 4. Determine the volume of the required substance, VCO , using the ideal gas law equation, PV = nRT. Remember that R = 8.314 kPa i L i mol PVCO = nCO RT 2 –1 –1 2 i K . 2 VCO = 2 = nCO RT 2 P (1.00 mol ) !" 8.314 kPa i L i mol –1 ( i K –1 # 273 K $ ) 101.325 kPa VCO = 22.4 L 2 Statement: When 5.60 L of butane undergoes complete combustion at STP, 22.4 L of carbon dioxide is produced. Copyright © 2011 Nelson Education Ltd. Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions 12.5-6 2. (a) Given: mass of potassium chlorate, mKClO = 122.6 g 3 molar mass of potassium chlorate, M KClO = 122.55 g/mol 3 The conditions are standard temperature and pressure, so T = 273 K and P = 101.325 kPa. Required: volume of oxygen, VO 2 Solution: Step 1. Write the balanced chemical equation, listing the given and required quantities. 2 KClO3 (s) ! 2 KCl(s) + 3 O 2 (g) 122.6 g VO 2 Step 2. Convert the mass of the given substance, mKClO , into an amount, nKClO . 3 nKClO = 122.6 g ! 3 3 1 mol 122.55 g nKClO = 1.000 mol 3 Step 3. Determine the amount of oxygen gas produced from the amount of potassium chlorate used. 3 molO 2 nO = 1.000 mol ! 2 2 molKClO 3 nO = 1.500 mol 2 Step 4. Determine the volume of the required substance, VO , using the ideal gas law equation, 2 PV = nRT. Remember that R = 8.314 kPa i L i mol–1 i K–1. PVO = nO RT 2 2 VO = 2 = nO RT 2 P (1.500 mol ) !" 8.314 kPa i L i mol –1 ( i K –1 # 273 K $ ) 101.325 kPa VO = 33.60 L 2 Statement: When 122.6 g of potassium chloride decomposes at STP, 33.60 L of oxygen gas is produced. Copyright © 2011 Nelson Education Ltd. Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions 12.5-7 (b) Given: mass of potassium chlorate, mKClO = 122.6 g 3 molar mass of potassium chlorate, M KClO = 122.55 g/mol 3 The conditions are standard temperature and pressure, so t = 25 °C and P = 100 kPa. Required: volume of oxygen, VO 2 Solution: Step 1. Convert the temperature value to kelvins. T = t + 273 = 25 + 273 T = 298 K Step 2. Write the balanced chemical equation, listing the given and required quantities. 2 KClO3 (s) ! 2 KCl(s) + 3 O 2 (g) 122.6 g VO 2 Step 3. Convert the mass of the given substance, mKClO , into an amount, nKClO . 3 nKClO = 122.6 g ! 3 3 1 mol 122.55 g nKClO = 1.000 mol 3 Step 4. Determine the amount of oxygen gas produced from the amount of potassium chlorate used. 3 molO 2 nO = 1.000 mol ! 2 2 molKClO 3 nO = 1.500 mol 2 Step 5. Determine the volume of the required substance, VO , using the ideal gas law equation, PV = nRT. Remember that R = 8.314 kPa i L i mol PVO = nO RT 2 –1 2 –1 i K . 2 VO = 2 = nO RT 2 P (1.500 mol ) !" 8.314 kPa i L i mol –1 ( i K –1 # 298 K $ ) 100 kPa VO = 37.16 L 2 Statement: When 122.6 g of potassium chloride decomposes at SATP, 37.2 L of oxygen gas is produced. Copyright © 2011 Nelson Education Ltd. Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions 12.5-8 3. Given: mass of sodium metal, mNa = 20.0 g molar mass of sodium metal, M Na = 22.99 g/mol temperature, t = 30 °C pressure, P = 125 kPa Required: volume of hydrogen gas, VH 2 Solution: Step 1. Convert the temperature value to kelvins. T = t + 273 = 30 + 273 T = 303 K Step 2. Write the balanced chemical equation, listing the given and required quantities. 2 Na(s) + 2 H 2O(g) ! H 2 (g) + 2 NaOH 20.0 g VH 2 Step 3. Convert the mass of the given substance, mNa , into an amount, nNa . nNa = 20.0 g ! 1 mol 22.99 g nNa = 0.870 mol Step 4. Determine the amount of hydrogen gas produced from the amount of sodium metal used. 1 molH 2 nH = 0.870 mol ! 2 2 molNa nH = 0.435 mol 2 Step 5. Determine the volume of the required substance, VH , using the ideal gas law equation, 2 PV = nRT. Remember that R = 8.314 kPa i L i mol–1 i K–1. PVH = nH RT 2 2 VH = 2 = nH RT 2 P (0.435 mol ) !" 8.314 kPa i L i mol –1 ( i K –1 # 303 K $ ) 125 kPa VH = 9 L 2 Statement: When 20.0 g of sodium metal reacts with water, 9 L of hydrogen gas is produced (under the given conditions of temperature and pressure). Copyright © 2011 Nelson Education Ltd. Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions 12.5-9 4. (a) Given: mass of octane, mC H = 64.2 kg 8 18 molar mass of octane, M C H = 114.26 g/mol 8 18 temperature, t = 80 °C pressure, P = 101.3 kPa Required: volume of carbon dioxide, VCO 2 Solution: Step 1. Convert the temperature value to kelvins. T = t + 273 = 80 + 273 T = 353 K Step 2. Write the balanced chemical equation, listing the given and required quantities. 2 C8 H18 (g) + 25 O 2 (g) ! 16 CO 2 (g) + 18 H 2O(g) 64.2 kg VCO 2 Step 3. Convert the mass of the given substance, mC H , into an amount, nC H . 8 18 8 18 mC H = 64.2 kg 8 18 = 64.2 kg ! 1000 g 1 kg mC H = 6.42 ! 104 g 8 18 nC H = 6.42 ! 104 g ! 8 18 nC H = 561.88 mol 8 18 1 mol 114.26 g [two extra digits carried] Step 4. Determine the amount of carbon dioxide produced from the amount of octane used. 16 molCO 2 nCO = 561.88 mol ! 2 2 molC H 8 nCO = 4495.0 mol 2 18 [two extra digits carried] Copyright © 2011 Nelson Education Ltd. Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions 12.5-10 Step 5. Determine the mass of carbon dioxide produced from the amount of carbon dioxide. M CO = 44.01 g/mol 2 mCO = nCO ! 2 2 44.01 g 1 kg ! 1 mol 103 g = 4495.0 mol ! 44.01 g 1 mol 1 kg = 1.98 ! 105 g ! 103 g = 1.98 ! 102 kg mCO = 2.0 ! 102 kg 2 Step 6. Determine the volume of the required substance, VCO , using the ideal gas law equation, PV = nRT. Remember that R = 8.314 kPa i L i mol–1 i K–1. PVCO = nCO RT 2 2 2 VCO = 2 = nCO RT 2 P (4495.0 mol ) !" 8.314 kPa i L i mol –1 ( i K –1 # 353 K $ ) 101.3 kPa VCO = 1.3 % 10 L 5 2 Statement: When a tank of 64.2 kg of gasoline undergoes combustion, 2.0 × 102 kg, or 1.3 × 105 L of carbon dioxide is produced (under the given conditions of temperature and pressure). (b) The mass of carbon dioxide produced from one tank of gas is 2.0 × 102 kg and each single mature tree can absorb 22 kg of carbon dioxide per year, therefore 2.0 ! 102 kg 22 kg =! 9 trees are required to absorb the carbon dioxide produced from one tank of gas in one year. Copyright © 2011 Nelson Education Ltd. Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions 12.5-11 5. (a) Given: mass of calcium carbonate, mCaCO = 1025 g 3 molar mass of calcium carbonate, M CaCO = 100.09 g/mol 3 temperature, t = 20 °C pressure, P = 100.0 kPa Required: volume of carbon dioxide, VCO 2 Solution: Step 1. Convert the temperature value to kelvins. T = t + 273 = 20 + 273 T = 293 K Step 2. Write the balanced chemical equation, listing the given and required quantities. CaCO3 (s) + 2 HCl(aq) ! CaCl2 (aq) + H 2O(l) + CO 2 (g) 1025 g VCO 2 Step 3. Convert the mass of the given substance, mCaCO , into an amount, nCaCO . 3 nCaCO = 1025 g ! 3 3 1 mol 100.09 g nCaCO = 10.24 mol 3 Step 4. Determine the amount of carbon dioxide produced from the amount of calcium carbonate used. 1 molCO 2 nCO = 10.24 mol ! 2 1 molCaCO 3 nCO = 10.24 mol 2 Step 5. Determine the volume of the required substance, VCO , using the ideal gas law equation, PV = nRT. Remember that R = 8.314 kPa i L i mol–1 i K–1. PVCO = nCO RT 2 2 2 VCO = 2 = nCO RT 2 P (10.24 mol ) !" 8.314 kPa i L i mol –1 ( i K –1 # 293 K $ ) 100.0 kPa VCO = 250 L 2 Statement: When 1025 g of calcium carbonate reacts with hydrochloric acid, 250 L of carbon dioxide is produced (under the given conditions of temperature and pressure). Copyright © 2011 Nelson Education Ltd. Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions 12.5-12 (b) Given: volume of carbon dioxide, VCO = 1550 L 2 temperature, T = 22.5 °C pressure, P = 100.0 kPa Required: mass of calcium carbonate, mCaCO 3 Solution: Step 1. Convert the temperature value to kelvins. T = t + 273 = 22.5 + 273 T = 295.5 K Step 2. Write the balanced chemical equation, listing the given and required quantities. CaCO3 (s) + 2 HCl(aq) ! CaCl2 (aq) + H 2O(l) + CO 2 (g) mCaCO 1550 L 3 Step 3. Determine the amount of the given substance, nCO , produced in the reaction using the 2 ideal gas law equation, PV = nRT. Remember that R = 8.314 kPa i L i mol–1 i K–1. PVCO = nCO RT 2 2 nCO = 2 = PVCO RT 2 (100.0 kPa )(1550 L ) (8.314 kPa i L i mol nCO = 63.091 mol 2 –1 )( i K –1 295.5 K ) [two extra digits carried] Step 4. Determine the amount of the calcium carbonate that is required to react with the amount of carbon dioxide. 1 molCaCO 3 nCaCO = 63.091 mol ! 3 1 molCO 2 nCaCO = 63.091 mol 3 [two extra digits carried] Copyright © 2011 Nelson Education Ltd. Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions 12.5-13 Step 5. Determine the mass of calcium carbonate that is required from the amount of calcium carbonate. M CaCO = 100.09 g/mol 3 mCaCO = nCaCO ! 3 3 100.09 g 1 kg ! 1 mol 103 g = 63.091 mol ! = 6.31 ! 103 g ! 100.09 g 1 mol ! 1 kg 103 g 1 kg 103 g mCaCO = 6.31 kg 3 Statement: 6.31 kg of calcium carbonate is required to produce 1550 L of carbon dioxide at these conditions. 6. Given: mass of ammonium sulfate, m( NH ) SO = 264.0 g 4 2 4 molar mass of ammonium sulfate, M ( NH ) SO = 132.17 g/mol 4 2 4 mass of potassium hydroxide, mKOH = 280.0 g molar mass of ammonium sulfate, M KOH = 56.11 g/mol temperature, t = 23 °C pressure, P = 64 kPa Required: volume of ammonia gas, VNH 3 Solution: Step 1. Convert the temperature value to kelvins. T = t + 273 = 23 + 273 T = 296 K Step 2. Write the balanced chemical equation, listing the given and required quantities. NH 4 2 SO 4 (aq) + 2 KOH(aq) ! 2 NH 3 (g) + K 2SO 4 (aq) + 2 H 2O(l) ( ) 264.0 g 280.0 g Copyright © 2011 Nelson Education Ltd. VNH 3 Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions 12.5-14 Step 3. Convert the masses of the given substances, m( NH ) SO and mKOH , into amounts, 4 2 4 n( NH ) SO and nKOH . 4 2 4 1 mol n( NH ) SO = 264.0 g ! 4 2 4 132.17 g n( NH ) SO = 1.997 mol 4 2 4 nKOH = 280.0 g ! 1 mol 56.11 g nKOH = 4.990 mol Since the mole ratio of ammonium sulfate to potassium hydroxide is 1 : 2, and the amount of ammonium sulfate is less than half of the amount of potassium hydroxide, the ammonium sulfate will limit the reaction and it is this amount that will be used to determine the amount of ammonia gas produced. Step 4. Determine the amount of ammonia gas produced from the amount of ammonium sulfate used. 2 molNH 3 nNH = 1.997 mol ! 3 1 mol( NH ) SO 4 2 4 nNH = 3.994 mol 3 Step 5. Determine the volume of the required substance, VNH , using the ideal gas law equation, PV = nRT. Remember that R = 8.314 kPa i L i mol–1 i K–1. PVNH = nNH RT 3 3 3 VNH = 3 = nNH RT 3 P (3.994 mol ) !" 8.314 kPa i L i mol –1 ( i K –1 # 296 K $ ) 64 kPa VNH = 150 L 3 Statement: When 264.0 g of ammonium sulfate reacts with 280.0 g of potassium hydroxide, 150 L of nitrogen dioxide is produced (under the given conditions of temperature and pressure). Copyright © 2011 Nelson Education Ltd. Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions 12.5-15 7. Given: mass of propane, mC H = 54.0 g 3 8 molar mass of propane, M C H = 44.11 g/mol 3 8 temperature, t = 25 °C pressure, P = 202.1 kPa Required: volume of carbon dioxide, VCO 2 Solution: Step 1. Convert the temperature value to kelvins. T = t + 273 = 25 + 273 T = 298 K Step 2. Write the balanced chemical equation, listing the given and required quantities. C3H 8 (g) + 5 O 2 (g) ! 3 CO 2 (g) + 4 H 2O(g) 54.0 g VCO 2 Step 3. Convert the mass of the given substance, mC H , into an amount, nC H . 3 nC H = 54.0 g ! 3 8 8 3 8 1 mol 44.01 g nC H = 1.2270 mol 3 8 [two extra digits carried] Step 4. Determine the amount of carbon dioxide produced from the amount of propane used. 3 molCO 2 nCO = 1.2270 mol ! 2 1 molC H 3 nCO = 3.6810 mol 2 8 [two extra digits carried] Step 5. Determine the volume of the required substance, VCO , using the ideal gas law equation, PV = nRT. Remember that R = 8.314 kPa i L i mol PVCO = nCO RT 2 –1 –1 2 i K . 2 VCO = 2 = nCO RT 2 P (3.6810 mol ) !" 8.314 kPa i L i mol –1 ( i K –1 # 298 K $ ) 202.1 kPa VCO = 45.1 L 2 Statement: When 54.0 g of propane reacts with excess oxygen, 45.1 L of carbon dioxide is produced (under the given conditions of temperature and pressure). Copyright © 2011 Nelson Education Ltd. Chapter 12: Gas Laws, Gas Mixtures, and Gas Reactions 12.5-16