標題:
Reacting Mass Quesion(F4)
發問:
13.70 g of an oxide of metal X (relative atomic mass = 207.0) reacts with excess hydrogen to produce 1.44 g of water.What is the empirical formula of the oxide?
最佳解答:
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2H2 + O2 --> 2H2O 1.44g water has 1.44/18 = 0.08 mole ∴Hydrogen is excess ∵Oxygen is the limited reactant By the formula, 2 mole of water is produced by 1 mole of oxygen ∵The mole of oxygen is 0.08/2 = 0.04 mole ∵The mass in the oxide is 0.04*32 = 1.28 g The remaining mass is the mass of the metal X = 13.7-1.28 = 12.42 g ∵The mole of the metal X is 12.42/207 = 0.06 mole By mole ratio The relative ratio of the oxygen is 0.04/0.04 = 1 The relative ratio of the metal X is 0.06/0.4 = 1.5 To make the number be integer, 1*2 = 2 1.5*2 = 3 . ∵The empirical formula of the oxide is X3O2 2007-04-19 10:39:54 補充： SOrry計mole ratio 果陣計錯左By mole ratioThe relative ratio of the oxygen atom is 0.08/0.06 = 1.3333The relative ratio of the metal X atom is 0.06/0.06 = 1To make the number be integer, 1.333*3 = 41*3 = 3∵The empirical formula of the oxide is X3O4
其他解答:
Molecular mass of H2O = 2x1 + 16 = 18 g mol-1 Fraction by mass of O in H2O = 16/18 Mass of O in 13.70 g of the oxide of X = Mass of O in 1.44 g of H2O = 1.44 x (16/18) = 1.28 g Mass of X in 13.70 g of the oxide of X = 13.70 - 1.28 = 12.42 g In the oxide of X, mole ratio X : O = (12.42/207) : (1.28/16) = 0.06 : 0.08 = 3 : 4 Empirical formula = X3O4 2007-04-18 22:30:49 補充： Checking :Suppose it is only known that 13.7 g of X3O4 is used.X3O4 4H2 → 3X 4H2ONo. of moles of X3O4 used = 13.7/685 = 0.02 molNo. of moles of H2O formed = 0.02 x 4 = 0.08 molMass of H2O formed = 0.08 x 18 = 1.44 gThis is consistent with the data given.
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