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標題:
mole concept question
發問:
in an experiment, two heated oxides X and Y of the same metal M were completely reduced by hydrogen. the results were recorded as follows: 1.mass of oxide X=42.9g,mass of metal M left=38.1g 2.mass of oxide Y=15.9g,mass of metal M left=12.7g if the formula of oxide X is M20 what is the formula of oxide Y?
最佳解答:
Denote the molar mass of M as y g mol-1. Molar mass of O = 16.0 g mol-1 Result 1: 42.9 g of M2O (X) contains 38.1 g of M. Fraction by mass of M in M2O = 2y/(2y 16) = 38.1/42.9 Hence, y = 63.5 The molar mass of M = 63.5 g mol-1 Result 2: 15.9 g of Y contains 12.7 g of M. Mass of M = 12.7 g Mass of O = 15.9 - 12.7 = 3.2 g Mole ratio M : O = (12.7/63.5) : (3.2/16) = 1 : 1 Empirical formula of Y = MO
其他解答:
mole concept question
發問:
in an experiment, two heated oxides X and Y of the same metal M were completely reduced by hydrogen. the results were recorded as follows: 1.mass of oxide X=42.9g,mass of metal M left=38.1g 2.mass of oxide Y=15.9g,mass of metal M left=12.7g if the formula of oxide X is M20 what is the formula of oxide Y?
最佳解答:
Denote the molar mass of M as y g mol-1. Molar mass of O = 16.0 g mol-1 Result 1: 42.9 g of M2O (X) contains 38.1 g of M. Fraction by mass of M in M2O = 2y/(2y 16) = 38.1/42.9 Hence, y = 63.5 The molar mass of M = 63.5 g mol-1 Result 2: 15.9 g of Y contains 12.7 g of M. Mass of M = 12.7 g Mass of O = 15.9 - 12.7 = 3.2 g Mole ratio M : O = (12.7/63.5) : (3.2/16) = 1 : 1 Empirical formula of Y = MO
其他解答:
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