標題:

chem實驗式(急~~)

發問:

金屬m的氧化物分解, 生出金屬m和氧氣. 分解後,氧化物的質量減少了6.9%, 氧化物的實驗式? (m的相對原子質量=107.9) 要step

 

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最佳解答:

There are 4 possibilities for the metal to form an oxide, depending on the valency of the metal. (1)M2O, (valency of the metal is one, in Group 1) 2M2O = 4M + O2 (2)MO, (valency of the metal is two, in Group II) 2MO = 2M + O2 (3)M2O3, (valency of the metal is three, in Group III) 2M2O3 = 4M + 3O2 (4)MO2, (valency of the metal is four, in Group IV) MO2 = M + O2 First, we consider case (1) where a metallic oxide M2O is formed (valency of M is 1) 2M2O = 4M + O2 Since the atomic mass of the metal is given = 107.9, Atomic mass of oxygen is 16 (From periodic table of elements or any chem. textbook) Molecular weight of M2O = (2 x 107.9 + 16) = 231.8 2M2O = 4M + O2 2(231.8) = 4(107.9) + 2 x (16) 463.6 = 431.6 + 32 463.6 g of M2O decomposes into 431.6 g of M and 32 g of oxygen The product, oxygen, is a gas which escapes into the atmosphere, leaving 431.6 g of metal M We try to find out percentage loss of oxygen with reference to the original metallic oxide 32 -------- x 100% = 6.9% 463.6 The answer is 6.9 % which agrees with the given percentage loss, so we know that empirical formula is M2O (two atoms of M combine with one atom of oxygen to form one molecule of M2O. ) If the answer is not 6.9% , you have to consider case 2, case 3 and case 4 in that order until the correct answer comes out. CORRECT ANSWER to the problem is M2O =================================================== Actually the metal is silver, Ag, with atomic mass = 107.9 2Ag2O = 4Ag + O2 I am sorry that the solution to your problem is in English as I cannot type in Chinese on the computer. I also cannot put the 2 into subscript as in M2O.

其他解答:D1B39E804036C6BD
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