chem problem(雜錦)－dkdqgav的部落格

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chem problem(雜錦)

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1. Naturally occurring argon(Ar) consist of 3 isotopes, their natural abundance and nuclear masses are as follow: 36Ar 0.337% 35.968 38Ar 0.0630% 37.963 40AR 99.600% 39.962Calculate the relative atomic mass of argon from these data2.An organic compound, Y,... 顯示更多 1. Naturally occurring argon(Ar) consist of 3 isotopes, their natural abundance and nuclear masses are as follow: 36Ar 0.337% 35.968 38Ar 0.0630% 37.963 40AR 99.600% 39.962 Calculate the relative atomic mass of argon from these data 2.An organic compound, Y, which contain only carbon,hydrogen and oxygen, has a molar mass of about 173g mol-1. When 0.430g of Y is burst in excess oxygen, 1100g of carbon dioxide and 0.450g of water are formed. (a) What is the empirical formula of Y (b) What is the molecular formula of Y 3. 1.00 mole of ethanoic acid is mixed with 1.00mole of ethanol at 25oC in a conical flask. After equilibrium is reached, the excess acid requires 66.6cm3 of 5.00M sodium hydroxide solution for complete neutralization. (a) Calculate the equilibrium constant for the esterification reaction. (b) In another conical flask,0.500 mole of ethanol is mixed with 1.00 mole of ethanoic acid at 25oC, predict the weight of ester formed at equilibrium.

最佳解答:

1. Relative atomic mass of Ar = (35.968*0.337 + 37.963*0.063 + 39.962*99.6)/100 = 39.947 2. (a) Molar mass of C = 12 g mol-1 Molar mass of H = 1 g mol-1 Molar mass of O = 16 g mol-1 Molar mass of H2O = 1*2 + 16 = 18 g mol-1 Molar mass of CO2 = 12 + 16 = 44 g mol-1 In 0.43 g of Y: Mass of C = Mass of C in CO2 = 1.100 * (12/44) = 0.3 g Mass of H = Mass of H in H2O = 0.450 * (2/18) = 0.05 g Mass of O = 0.43 - (0.3 + 0.05) = 0.08 g Mole ratio C : H : O = 0.3/12 : 0.05/1 : 0.08/16 = 0.025 : 0.05 : 0.005 = 5 : 10 : 1 Empirical formula = C5H10O (b) Let the molecular formula be (C5H10O)n n(5*12 + 10*1 + 16) ≒ 173 n = 2 Molecular formula = C10H20O2 3. (a) NaOH + CH3COOH → CH3COONa + H2O No. of moles of NaOH in titration = 5 * (66.6/1000) = 0.333 mol No. of moles of CH3COOH at equilibrium = 0.333 mol CH3COOH + CH3CH2OH = CH3COOCH2CH3 + H2O Let the volume of the reaction mixture be V dm3. At equilibrium: [CH3COOH] = [CH3CH2OH] = 0.333/V mol dm-3 [CH3COOCH2-CH3] = [H2O] = 0.667/V mol dm-3 Equilibrium constant, Kc = [CH3COOCH2-CH3]*[H2O] / [CH3COOH]*[CH3CH2OH] = (0.667/V)2/(0.333/V)2 = 4.01 (b) CH3COOH + CH3CH2OH = CH3COOCH2CH3 + H2O Let the amount of CH3COOCH2CH3 at equilibrium be y mol, and let the volume of the system be V' dm3. At equilibrium: [CH3COOCH2CH3] = [H2O] = y/V' mol dm-3 [CH3COOH] = (1 - y)/V' mol dm-3 [CH3CH2OH] = (0.5 - y)/V' mol dm-3 Kc = [CH3COOCH2-CH3]*[H2O] / [CH3COOH]*[CH3CH2OH] 4.01 = (y/V')2/[(1 - y)/V']*[(0.5 - y)/V'] 4.01 = y2/(0.5 - 1.5y + y2) 2.005 - 6.015y + 4.01y2 = y2 3.01y2 - 6.015y + 2.005 = 0 y = 0.423 ooro y = 1.07 (rejected) Molar mass of CH3COOCH2CH3 = 12*4 + 8*1 + 16*2 = 88 g mol-1 Mass of ester formed = 0.423 * 88 = 37.2 g =

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