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8 Αυγ 2015 · In aqueous solution $\ce{Fe2(SO4)3}$ dissociates into $\ce{Fe^{3+} (aq)}$ and $\ce{SO4^{2-} (aq)}$ ions. The sulphate ions will be solvated by hydrogen bonding with the water and the iron ions will form the hexaaquairon(III) complex, $\ce{[Fe(H2O)6]^{3+}}$.
11 Οκτ 2015 · $$\ce{2Fe^2+ + SO4^2- + 4H+ -> 2Fe^3+ + SO2 + 2H2O}$$ In the dilute acid, this equilibrium lies far to the left because the concentrations of $\ce{SO4^2-}$ and $\ce{H+}$ are rather low. However, making the acid more concentrated shifts the equilibrium to the right so that you get some of the products.
11 Μαΐ 2015 · As @Greg pointed out, oxygen appears to be acting as the oxidiser here and it forms $\ce{Fe2O3}$. There is not enough sulphate present in the starting materials to form only $\ce{Fe2(SO4)3}$ and hence the oxide is formed as well.
11 Σεπ 2015 · $$\ce{KMnO4 + FeSO4 + H2SO4 -> K2SO4 + MnSO4 + Fe2(SO4)3 + H2O}$$ I found out that manganese is reduced as the oxidation number goes from $+7$ to $+2$ and iron gets oxidized as the oxidation number goes from $+2$ to $+3$. I wrote the two half reactions, \eqref{Q:red} for the reduction and \eqref{Q:ox} for the oxidation. Did I get them right?
7 Ιουν 2021 · In this equation, the iron atom is reduced from + $3$ to + $2$. This is impossible in an oxidation with $\ce{O2}$. The oxidative action of $\ce{O2}$ cannot lead to a decrease of the oxidation number of an atom present in the reaction. The oxidation of $\ce{Fe2S3}$ must produce a ferric compound, like $\ce{Fe2(SO4)3}$ instead.
In contrast, if $\ce{Fe2(SO4)3}$ were used, there would only be aqueous ions. At the copper surface, there is only a one-electron oxidation of $\ce{Cu}$ to $\ce{Cu+}$ . As the above article explains, without coordination of chloride, the $\ce{Cu+}$ would be essentially insoluble.
It is clear that the reaction (Fe2) releases more energy and thus should be preferred over the other reaction. This means that we expect a yield of 100% Cu and not 150% Cu. Now that we have gathered some theoretical understanding of the matter, let's proceed to the interesting part, which is
26 Φεβ 2015 · Wikipedia tells me that the production of iron (III) sulfate is: 2FeSOX4 +HX2SOX4 +HX2OX2 FeX2(SOX4)X3 +2HX2O 2 F e S O X 4 + H X 2 S O X 4 + H X 2 O X 2 F e X 2 (S O X 4) X 3 + 2 H X 2 O. I was wondering if there was a difference between salt formation equation and production, because I thought it would've went something more like: 2Fe(OH)X3 ...
14 Ιουν 2018 · $$\ce{k4fe(cn)6 + kmno4 + h2so4 + h2so4 -> khso4 + fe2(so4)3 + mnso4 + hno3 + co2 + h2o}$$ My solution: $$\ce{a K4Fe(CN)6 + b KMnO4 + c H2SO4 -> d KHSO4 + e Fe2(SO4)3 + f MnSO4 + g HNO3 + h CO2 + i H2O}$$
The molar mass of $\ce{Fe2(SO4)3}$ is $399.91 \, \text{g}/\text{mol}$. How would you go about solving a question like that? I'm confused about how to use only the quantity of sulfate ions to find the mass in grams of the whole sample, given it's grams per mole.
