定义组件的矿物组合?- 江南体育网页版- - - - -地球科学堆江南电子竞技平台栈交换 最近30从www.hoelymoley.com 2023 - 07 - 07 - t14:11:48z //www.hoelymoley.com/feeds/question/6621 https://creativecommons.org/licenses/by-sa/4.0/rdf //www.hoelymoley.com/q/6621 6 定义组件的矿物组合? 墨丘利- 197 //www.hoelymoley.com/users/4479 2015 - 09 - 30 - t21:16:56z 2015 - 10 - 04 - t19:19:53z < p >由于矿物的矿物组合$ $ cp (\ ce {CuFeS_2}), bn (\ ce {Cu_5FeS_4})、cc (\ ce {Cu_2S}),太(\ ce {Fe_3O_4}), py (\ ce {FeS_2}) $ $ cp, bn,太接触,和py, cp, bn, cc接触。什么是组件?< / p > < p >我被告知# = #成分- #反应-组件#约束但我不知道如何定义成分或约束。我不知道反应的数量。我试图定义成分基于金属的氧化态(假设和O总是2),和我$ $ \ ce{铜^ {+ 1},Fe ^ {+ 4}, Fe ^ {+ 3}, Fe ^ {+ 2}} $ $ $ $ \ ce的{Fe ^ {+ 3}}: cp (\ ce {CuFeS_2}), bn (\ ce {Cu_5FeS_4}) $ $和$ $ \ ce {Fe ^ {+ 4}}: py (\ ce {FeS_2}) $ $和$ $ \ ce {Fe ^ {+ 3}, Fe ^{+ 2}}:太(\ ce {Fe_3O_4}) $ $ < / p > < p >对于反应的数量,我很困惑,因为只有一个矿物的氧(因此不能反应的一部分的其他人产生另一个),和我之间很难平衡反应。< / p > < p >谁能给我什么建议如何算出组件,一般来说我应该如何定义组件、成分、基于组合的矿物质和反应?

This question is based on a problem from Walther "Essentials of Geochemistry" here is a [Link] to the question.

Just to clarify based on the given answers, the components are cc, cp, py, and mt? Since bn is the only one that can be formed from a reactions of the others?

 //www.hoelymoley.com/questions/6621/-/6635 # 6635 2 回答Gimelist定义组件的矿物组合? Gimelist //www.hoelymoley.com/users/725 2015 - 10 - 02 - t11:18:58z 2015 - 10 - 02 - t11:18:58z < p >让我们开始了解什么是组件。系统的组件所需的最少的化学“物种”来描述一个系统。这是什么意思?我们来看一个大家熟悉的例子。方镁石、镁橄榄石、顽辉石和石英。分别以,Mg 2 <子> < /订阅> SiO <子> 4 < /订阅>,MgSiO <子> 3 < /订阅>和SiO 2 <子> < /订阅>。所以我们可以说我们有4个组件,每一个公式我只是指出。但这并不是最低数量。我们可以得到一个更小的数字。所以我们可以说,例如,我们的组件是毫克,Si和o .我们可以创建所有的矿物公式从这三个组件。 But wait - we can go even lower! If we choose our components to be only MgO and SiO2, we have 2 components that can still describe our entire system. We can combine 2MgO+1SiO2 to describe forsterite and 1MgO+1SiO2 to describe enstatite. Basically, the components have to be independent. If there is one component that can be described by others, then it is not actually a component.

So let's go to your problem. First of all, I would say that your assumption that the oxidation state of Cu is +1 and that of S is -2 is not entirely correct. In pyrite, S is actually -1 and Fe is +2 (Fe4+ does not exist). As an aside, the oxidation state of Cu (and others) in sulfides is a can of worms: those are not ionic compounds but rather a mixture of metallic and covalent compounds where oxidation state does not mean much, especially in the more complex cases such as bornite and chalcopyrite.

That said, if you look carefully and try to balance your reactions (try to make one mineral using the others), you can see that bn = cp + 2cc, so bn is not a component. Can you still reduce the amount of components? No. Fe3O4 must be there, and there is no way you can make CuFeS2 from Cu2S and FeS2. You might think about using the components S-Fe-Cu-O, but that would actually be incorrect because for that to happen S has to be the same valence, which obviously is not (compare Cu2S2- and FeS1-2).
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