Chemistry - Calculations－lgzrelv的部落格

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Chemistry - Calculations

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Pleae help me solve the following questions :1. Complete combustion of 1.86g of an organic compound gave 2.64g of carbon dioxide and 1.62g of water as the only products. If the density of the compound at s.t.p. was found to be 2.77 g/dm^3, find its molecular formula.2. A compound Z contained only... 顯示更多 Pleae help me solve the following questions : 1. Complete combustion of 1.86g of an organic compound gave 2.64g of carbon dioxide and 1.62g of water as the only products. If the density of the compound at s.t.p. was found to be 2.77 g/dm^3, find its molecular formula. 2. A compound Z contained only nitrogen, hydrogen and chlorine. On quantitative decomposition in aqueous solution, 3.17g of Z liberated 1109cm^3 of nitrogen at room conditions. Addition of aqueous silver nitrate to a solution containing 1.585g of Z caused the formation of 3.315g of silver chloride. Deduce the simplest formula of Z. 3. Molar aqueous solutions of BaCl2 and a metal sulphate of molar mass 342g were prepared. (a). What mass of anhydrous BaCl2 would be needed to prepare 250cm^3 of a molar solution? (b). A white precipitate was formed when the two solutions were mixed. What is this white precipitate?

最佳解答:

(1) CxHyOz + ? O2 → x CO2 + y/2 H2O no. of mole of CO2 formed = 2.64 / (12.0 + 16.0 x 2) = 0.06 mol no. of mole of H2O formed = 1.62 / (1.0 x 2 + 16.0) = 0.09 mol mass of O in the organic compound = 1.86 - (0.06 x 12.0 + 0.09 x 2 x 1.0) = 0.96 g no. of mole of O in the organic compound = 0.96 / 16 = 0.06 mol no. of mole of C : no. of mole of H : no. of mole of O = 0.06 : 0.18 : 0.06 = 1 : 3 : 1 empirical formula is CH3O PV = nRT PM = dRT (M = molar mass, d = density) M = 2.77 x 0.0821 x 273 M = 62.1 g/mol 62.1 = (12.0 + 1.0 x 3 + 16.0) x x = 2 molecular formula is C2H6O2 (2) no. of mole of N2 formed = 1.109 / 24 = 0.0462 mol no. of mole of N in 3.17 g of compound Z = 0.0462 x 2 = 0.0924 mol Ag+(aq) + Cl-(aq) → AgCl(s) no. of mole of AgCl = 3.315 / (107.9 + 35.5) = 0.0231 mol no. of mole of Cl in 3.17 g of compound Z = 0.0231 x 2 = 0.0462 mol no. of mole of H in 3.17 g of compound Z = (3.17 - 0.0462 x 35.5 - 0.0924 x 14.0) / 1.0 = 0.2363 mol no. of mole of N : no. of mole of H : no. of mole of Cl = 0.0924 : 0.2363 : 0.0462 = 2 : 5 : 1 empirical formula is NH5Cl (3a) no. of mole of BaCl2 in a 250 cm^3 solution = 1 x 0.25 = 0.25 mol mass of 0.25 mol of BaCl2 = 0.25 x (137.3 + 35.5) = 43.2 g (3b) BaSO4

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