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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5.97You can map out the following strategy to solve for the total volume of gas. grams nitroglycerin moles nitroglycerin moles products volume of products 21 mol nitroglycerin29 mol product? mol products2.610g nitroglycerin8.3 mol227.09 g nitroglycerin4 mol nitroglycerin=×××=Calculating the volume of products: productL atm(8.3 mol) 0.0821(298 K)mol K(1.2 atm)===2product1.710 LnRTPV×The relationship between partial pressure and Ptotalis: Pi=ΧiPTCalculate the mole fraction of each gaseous product, then calculate its partial pressure using the equation above. componentmoles componenttotal moles all components=Χ
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CHAPTER 5: GASES 15822CO12 mol CO0.4129 mol product==ΧSimilarly, 2H OΧ=0.34, 2NΧ=0.21, and 2OΧ=0.034 22COCOT=PPΧ(0.41)(1.2 atm)==2CO0.49 atmPSimilarly, ,, and.===222H ONO0.41 atm0.25 atm0.041 atmPPP5.98We 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. 30.001 L(0.74 atm) 97.2 mL1 mL1.8510molL atm0.0821(200273)Kmol K-×===×+PVnRT30.145 gmolar mass78.4 g/mol1.8510mol-==×The empirical mass of CH =13.02 g/mol Since 78.4 g/mol6.02613.02 g/mol=, the molecular formula is (CH)6or C6H6. 5.99(a) NH4NO2(s) →N2(g) +2H2O(l) (b) Map out the following strategy to solve the problem. volume N2moles N2moles NH4NO2grams NH4NO2First, calculate the moles of N2using the ideal gas equation. T(K) =22°+273°=295 K 1 L86.2 mL0.0862 L1000 mL=×=V22NN=PVnRT23N(1.20 atm)(0.0862 L)=4.2710molL atm0.0821(295 K)mol K-=×nNext, calculate the mass of NH4NO2needed to produce 4.27 ×10-3mole of N2. 3424222421 mol NH NO64.05 g NH NO(4.2710mol N )1 mol N1 mol NH NO-=×××=42? g NH NO0.273 g
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CHAPTER 5: GASES 1595.100The reaction is: HCO3-(aq) +H+(aq) →H2O(l) +CO2(g) The mass of HCO3-reacted is: 3332.5% HCO3.29 g tablet1.07 g HCO100% tablet--×=32232331 mol HCO1 mol COmol COproduced1.07 g HCO0.0175 mol CO61.02 g HCO1 mol HCO----=××=22COL atm(0.0175 mol CO ) 0.0821(37273)Kmol K0.445 L(1.00 atm)+====2CO445 mLnRTPV5.101No, 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)4formed is: 441 mol Ni(CO)1 mol Ni86.4 g Ni1.47 mol Ni(CO)58.69 g Ni1 mol Ni××=The pressure of Ni(CO)4is: L atm(1.47 mol) 0.0821(43 + 273)Kmol K4.00 L===9.53 atmnRTVP(b)Ni(CO)4decomposes to produce more moles of gas (CO), which increases the pressure. Ni(CO)4(g) →Ni(s) +4CO(g) 5.103The partial pressure of carbon dioxide is higher in the winter because carbon dioxide is utilized less by photosynthesis in plants. 5.104Using the ideal gas equation, we can calculate the moles of gas.
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