化學問題-求過度金屬－hzb53jl55v的部落格

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化學問題-求過度金屬

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一個25g的樣本,系第一行的過度金屬(元素21到30,也就繫Sc,Ti,V,Cr,Mn,Fe,Co,Ni,Cu,Zn)其中一個, 系氧氣中加熱,生成一個獨立純淨化合物, 重量系44.63g.,確認過度金屬你可以設想果個過度金屬生成一個離子化合物,也就繫某金屬含有整數charge(+1,+2,+3...).同時.生成氧氣含有O2-). 求樣本用的系咩過度金屬? (麻煩冩解答過程,THX)

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Transition Metal 3d 21 Sc Scandium 44.959 g/mol 22 Ti Titanium 47.9 g/mol 23 V Vanadium 50.9415 g/mol 24 Cr Crromium 51.996 g/mol 25 Mn Manganese 54.938 g/mol 26 Fe Iron 55.847 g/mol 27 Co Cobalt 58.9332 g/mol 28 Ni Nickel 58.71 g/mol 29 Cu Copper 63.546 g/mol 30 Zn Zinc 65.38 g/mol 8 O Oxygen 15.9994 g/mol Let Mt be the required transition metal, k be the number of oxyen atom per metal atom in the resulted oxide. Mt + (k) * O-- --> Mt O(k) + 2(k) * e- resulted weight of oxide is 25 + (15.9994 / ( Mt / 25)) g First we try to determine the possible oxidation states for the metal If the original metal is 25g, then if the metal is Mt(+I), the oxide is Mt2O, the range of weight will be in the range Sc 29.45g to Zn 28.06g which is not the required range if the metal is Mt(+II), the oxide is MtO, the range of weight will be in the range Sc 33.90g to Zn 31.12g which is not the required range if the metal is Mt(+III), the oxide is Mt2O3, the range of weight will be in the range Sc 38.34g to Zn 34.18g which is not the required range if the metal is Mt(+IV), the oxide is MtO2, the range of weight will be in the range Sc 42.79g to Zn 37.24g which is not the required range if the metal is Mt(+V), the oxide is Mt2O5, the range of weight will be in the range Sc 47.24g to Zn 40.29g which is the required range if the metal is Mt(+VI), the oxide is MtO3, the range of weight will be in the range Sc 51.69g to Zn 43.35g which is the required range if the metal is Mt(+VII), the oxide is Mt2O7, the range of weight will be in the range Sc 56.14g to Zn 46.41g which is not the required range so the metal will only be Mt(+V) or Mt(+VI) as the resulted oxide weight is 44.63g From the characteristic of those transition metal, we know that their oxidation state is related to their electronic structures. For those ten transition metals, common electronic structure 1s2 2s2 2p6 3s2 3p6 21 Sc 3d1 4s2 22 Ti 3d2 4s2 23 V 3d3 4s2 24 Cr 3d5 4s1 25 Mn 3d5 4s2 26 Fe 3d6 4s2 27 Co 3d7 4s2 28 Ni 3d8 4s2 29 Cu 3d10 4s1 30 Zn 3d10 4s2 For oxidation states of +V or +VI to exist, the required metal can only be 23 V / 24 Cr / 25 Mn / 26 Fe / 27 Co. For oxidation states of +V, weight of metal oxide (Mt2O5) with 25g of metal is 21 Sc 47.24g 22 Ti 45.88g 23 V 44.63g 24 Cr 44.23g 25 Mn 43.20g 26 Fe 42.91g 27 Co 41.97g 28 Ni 42.03g 29 Cu 40.74g 30 Zn 40.29g For oxidation states of +VI, weight of metal oxide (MtO3) with 25g of metal is 21 Sc 51.69g 22 Ti 50.05g 23 V 48.56g 24 Cr 48.08g 25 Mn 46.84g 26 Fe 46.49g 27 Co 45.36g 28 Ni 45.44g 29 Cu 43.88g 30 Zn 43.35g After excluding of Sc, Ti, Ni, Cu, and Zn, we get For oxidation states of +V, oxide weight 23 V 44.63g 24 Cr 44.23g 25 Mn 43.20g 26 Fe 42.91g 27 Co 41.97g For oxidation states of +VI, oxide weight 23 V 48.56g 24 Cr 48.08g 25 Mn 46.84g 26 Fe 46.49g 27 Co 45.36g As a result, the metal is 23 V Vanadium, with oxidation states of +V form oxide V2O5.

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