597 You can map out the following strategy to solve for the total volume of gas

# 597 you can map out the following strategy to solve

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5.97 You can map out the following strategy to solve for the total volume of gas. grams nitroglycerin moles nitroglycerin moles products volume of products 2 1 mol nitroglycerin 29 mol product ? mol products 2.6 10 g nitroglycerin 8.3 mol 227.09 g nitroglycerin 4 mol nitroglycerin = × × × = Calculating the volume of products: product L atm (8.3 mol) 0.0821 (298 K) mol K (1.2 atm) × ÷ × = = = 2 product 1.7 10 L n RT P V × The relationship between partial pressure and P total is: P i = Χ i P T Calculate the mole fraction of each gaseous product, then calculate its partial pressure using the equation above. 165
CHAPTER 5: GASES component moles component total moles all components = Χ 166
CHAPTER 5: GASES 2 2 CO 12 mol CO 0.41 29 mol product = = Χ Similarly, 2 H O Χ = 0.34, 2 N Χ = 0.21, and 2 O Χ = 0.034 2 2 CO CO T = P P Χ (0.41)(1.2 atm) = = 2 CO 0.49 atm P Similarly, , , and . = = = 2 2 2 H O N O 0.41 atm 0.25 atm 0.041 atm P P P 5.98 We need to determine the molar mass of the gas. Comparing the molar mass to the empirical mass will allow us to determine the molecular formula. 3 0.001 L (0.74 atm) 97.2 mL 1 mL 1.85 10 mol L atm 0.0821 (200 273)K mol K - × ÷ = = = × × + ÷ × PV n RT 3 0.145 g molar mass 78.4 g/mol 1.85 10 mol - = = × The empirical mass of CH = 13.02 g/mol Since 78.4 g/mol 6.02 6 13.02 g/mol = , the molecular formula is (CH) 6 or C 6 H 6 . 5.99 (a) NH 4 NO 2 ( s ) → N 2 ( g ) + 2H 2 O( l ) (b) Map out the following strategy to solve the problem. volume N 2 moles N 2 moles NH 4 NO 2 grams NH 4 NO 2 First, calculate the moles of N 2 using the ideal gas equation. T (K) = 22 ° + 273 ° = 295 K 1 L 86.2 mL 0.0862 L 1000 mL = × = V 2 2 N N = P V n RT 2 3 N (1.20 atm)(0.0862 L) = 4.27 10 mol L atm 0.0821 (295 K) mol K - = × × ÷ × n Next, calculate the mass of NH 4 NO 2 needed to produce 4.27 × 10 - 3 mole of N 2 . 3 4 2 4 2 2 2 4 2 1 mol NH NO 64.05 g NH NO (4.27 10 mol N ) 1 mol N 1 mol NH NO - = × × × = 4 2 ? g NH NO 0.273 g 167
CHAPTER 5: GASES 5.100 The reaction is: HCO 3 - ( aq ) + H + ( aq ) → H 2 O( l ) + CO 2 ( g ) The mass of HCO 3 - reacted is: 3 3 32.5% HCO 3.29 g tablet 1.07 g HCO 100% tablet - - × = 3 2 2 3 2 3 3 1 mol HCO 1 mol CO mol CO produced 1.07 g HCO 0.0175 mol CO 61.02 g HCO 1 mol HCO - - - - = × × = 2 2 CO L atm (0.0175 mol CO ) 0.0821 (37 273)K mol K 0.445 L (1.00 atm) × + ÷ × = = = = 2 CO 445 mL n RT P V 5.101 No, because an ideal gas cannot be liquefied, since the assumption is that there are no intermolecular forces in an ideal gas. 5.102 (a) The number of moles of Ni(CO) 4 formed is: 4 4 1 mol Ni(CO) 1 mol Ni 86.4 g Ni 1.47 mol Ni(CO) 58.69 g Ni 1 mol Ni × × = The pressure of Ni(CO) 4 is: L atm (1.47 mol) 0.0821 (43 + 273)K mol K 4.00 L × ÷ × = = = 9.53 atm nRT V P (b) Ni(CO) 4 decomposes to produce more moles of gas (CO), which increases the pressure. Ni(CO) 4 ( g ) → Ni( s ) + 4CO( g ) 5.103 The partial pressure of carbon dioxide is higher in the winter because carbon dioxide is utilized less by photosynthesis in plants. 5.104 Using the ideal gas equation, we can calculate the moles of gas.