ctg202 Posted November 22, 2003 Share Posted November 22, 2003 1)KMnO4+H2SO4+Na2SO3 -> Mn+2+SO4-2 2)KMnO4+NaOH+Na2SO3->Mn(OH)2+SO4-2 3)K2C2O4+H2SO4+Na2SO3->Cr+3 +SO4-2 4)K2C2O4+NaOH+ Na2SO3->CO4-2+SO4-2 5)K2CrO4+H2SO4+Na2SO3->Cr+3+SO4-2 6)K2CrO4+NaOH+Na2SO3->Cr2O7-2+SO4-2 Does anyone know how to balance these equations? I know you have to add some things from the left side to the right but i dont know what. Please help! Link to comment Share on other sites More sharing options...
NSX Posted November 22, 2003 Share Posted November 22, 2003 We did something like this in Gen. Chem.; balancing Redox RXs in basic or acidic sol'n. But your examples are kind of different. I don't see where the Na's come from... Well, w/ the Redox RXs, you add H2O, H+, and OH- accordingly, but I'm not sure how to apply it for your examples... Sorry. Link to comment Share on other sites More sharing options...
Atlantic Posted November 22, 2003 Share Posted November 22, 2003 you'll have to figure out the charge of every part of those equations (i.e. polyatomic ions) , and changes their quantities to make the balences equal for both sides. Link to comment Share on other sites More sharing options...
ctg202 Posted November 23, 2003 Author Share Posted November 23, 2003 The +4's and other numbers are the different charges, I just didn't know how to write a superscript. The number after each equation is supposed to be a subscript. Thanks for trying everyone. Link to comment Share on other sites More sharing options...
YT2095 Posted November 23, 2003 Share Posted November 23, 2003 Na2SO3? where`s that involved at all? and why is in each one? sodium sulphate is practicaly inert in these equasions!? do me a fave, present them one at a time (or show your source) and use normal chem equasions, as non of that seems to make any sense at all? take for example number #1 potassium permanganate + sulphuric acid + sodium sulphate (for a start what on earth is the sodium sulphate in there for?) I`de haxxard to say at 1`st glance it wouldn`1t react at all, but over time MAY liberate a few O2 bubbles (at 1`st glance). it needs tidying up ps: I like your choice of Oxidiser in #5 and 6 though Link to comment Share on other sites More sharing options...
ctg202 Posted November 23, 2003 Author Share Posted November 23, 2003 I dont know why the elements that are in there are in there. It was a chemistry lab in my class. We were to mix the first things in each of the equations, and then it would turn a certain color. In the second one for example, it turned into a dark brown solid and that supposedly meant it turned into what you see on the other side of the arrow in each equation. Im not sure if i am setting up the equations correctly however. It was an expirement on oxidation reduction. I hope that helps. Thanks again for looking. Link to comment Share on other sites More sharing options...
YT2095 Posted November 23, 2003 Share Posted November 23, 2003 well eqasion #1 has no reducing agents in it? potassium permanganate and sulphuric acid are both considered oxidising, your sodium sulphate would be considered neutral. was "Displacement" mentioned at all? l Link to comment Share on other sites More sharing options...
ctg202 Posted November 23, 2003 Author Share Posted November 23, 2003 I don't remember hearing anything about displacement. Oh well, I guess my equations make no sense. Thanks everybody. Link to comment Share on other sites More sharing options...
YT2095 Posted November 23, 2003 Share Posted November 23, 2003 they DO, in the sense that each chem is correctly identifiable. the problem comes from above stated reasons maybe? (just a guess) the other useless stuff in it is there deliberately to try and confuse you, if that`s the case, just factor them out and re-write them as I said, they need to be more tidy Link to comment Share on other sites More sharing options...
fafalone Posted November 23, 2003 Share Posted November 23, 2003 A better question than where the Na comes from is where does it go in all of them.. I'm guessing you'll form Na-A, where A is the anion in the products (negatively charged ion). To balance it, you just have to make equal numbers of atoms on both sides... if there's 2Cr's on one side, then there should be 2 on the other side. It's not hard, just takes practice. There's programs that will do it for you, but I'd strongly recommend against using them, as in more advanced chemistry courses you'll need to balance like its 2nd nature. Link to comment Share on other sites More sharing options...
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