Balance the reaction: MnS + As[sub]2[/sub]Cr[sub]10[/sub]O[sub]35[/sub] + H[sub]2[/sub]SO[sub]4[/sub] -> HMnO[sub]4[/sub] + AsH[sub]3[/sub] + CrS[sub]3[/sub]O[sub]12[/sub] + H[sub]2[/sub]O It's a curly one, but there's a relatively easy way to solve it. Suggestions? This problem is a question in a linear algebra assignment, and yes, I know the answer.
Start with Manganese, Arsenic, and Chromium, which are each in only one compound on each side. You can then write three equations (for Hydrogen, Sulfur, and Oxygen) in four unknowns, and solve for the lowest integer solution. The equations are: x + z = 30y 35y + 4z = 4x + 120y + w 2z = 6y + x + 2w ...which simplify to... 16y = 13x 16z = 374x 16w = 327x ...and the answer follows trivially.
Pete, could you identify what x,y,z are? It would be a bit easier to follow. What we were "supposed" to do was write six equations in 7 unknowns; you get a system where each row corresponds to an element and each column corresponds to a species. Then you just row reduce it.
I didn't want to do all the work for you Please Register or Log in to view the hidden image! x is MnS y is As2Cr10O35 z is H2SO4 w is H2O Yep, that's pretty much the same as what I did, except that I first removed the trivial equations for Manganese, Arsenic and Chromium . For completeness, the remaining equations are: v = x u = 2y t = 10y Where... v is HMnO4 u is AsH3 t is CrS3O12 But these are not the same compounds as in arfa brane's problem?
This is balanced 4MnS + As2(Cr2O7)5 + H2SO4 ----> 40 e +2AsH3 + 5Cr2(SO4)3 +. 10 H2O MnS + As2Cr10O35 + H2SO4 ------> AsH3 + CrS3O12 + 12 H2O
Hmm. Cr2(SO4)3 has one more Chromium than CrS3O12 But, CrS3O12 doesn't seem to be an actual compound, so now I don't know!
Chromium is reduced from +6 Cr to +3, and Sulfur is oxidized from -2 to + 4 10 x 3 =30 e 4 x 6 + 24 e Arsenic is reduced from + 5 to As 0 Manganese is oxidized from +2 to + 6 2 x 5 = 10e 4 x 4 + 16 e ----------------------------------------------------------------------------------- 40 e 40 e
I don't think the mathematical problem is concerned about whether certain combinations of elements make sense--for instance HMnO[sub]4[/sub] is missing an H, the stable compound is H[sub]2[/sub]MnO[sub]4[/sub]; maybe the reaction products are transitional? Can't find any reference to CrS[sub]3[/sub]O[sub]12[/sub], but Cr looks like it's hexavalent. Anyway, the mathematical solution is to write down the six equations, one for each element, and use row reduction. Then you just parametrise the last column variable (which is the integral # of water molecules) once you have row reduced echelon form. OR is that reduced row echelon form? No matter, you end up with single 1's in each column except for the last.
I don't understand. My chemistry isn't that hot any more. All I know is that in arfra brane's has CrS3O12 in his problem, you have Cr2(SO4)3 in your solution, and that these are not the same thing. Are you saying that CrS3O12 is a typo, and the problem should have had Cr2(SO4)3?
Agreed (except I think HMnO[sub]4[/sub] is fine. Permanganate ion has a charge of neg one?). arauca's Sulfur isn't balanced anyway. I did it by kludgy simultaneous solutions instead of matrix reduction, but the effect should be identical. I got some surprisingly large numbers in the end, but it seems to check out: 16 MnS + 13 As[sub]2[/sub]Cr[sub]10[/sub]O[sub]35[/sub] + 374 H[sub]2[/sub]SO[sub]4[/sub] -> 16 HMnO[sub]4[/sub] + 26 AsH[sub]3[/sub] + 130 CrS[sub]3[/sub]O[sub]12[/sub] + 327 H[sub]2[/sub]O Changing CrS[sub]3[/sub]O[sub]12[/sub] to Cr[sub]2[/sub]S[sub]3[/sub]O[sub]12[/sub] doesn't make it any nicer.
I believe for balancing Cr(SO4)3 would not mach and even the Cr have a valance +2 +3 +6 the 2 valance will not satisfy, +3 will satisfy so it will be Cr2 (SO4)3 and because we start with 10 chrome atoms at the end there should be 10 Chromium atoms so 5Cr2(SO4)3 and satisfy the balance of sulfur from Manganese sulfide. At the end the Oxidation Reduction reaction in electron transfer have to be satisfied. and this is what I am attempting to show.