rainingspiders Posted October 26, 2014 Share Posted October 26, 2014 I'm supposed to balance the following equation: [latex] HCl + As2O3 + NaNO3 + H2O -> NO + H3AsO4 + NaCl [/latex] Anyways, I got stuck stuck here... [latex] HCl + As2O3 + NaNO3 + 3H2O -> NO + 2H3AsO4 + NaCl [/latex] No matter what I do I couldn't get H and O to balance out at the same time! When I balance O, the H is out of balance, and when I balance H, O is out of balance! Augh! How do I do this? Ok, I have solved it to here... [latex] 2HCl + As2O3 + 2NaNO3 + 2H2O -> 2NO + 2H3AsO4 + 2NaCl [/latex] There are 11 O on the right and 10 O on the left. ... Now what? Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted October 26, 2014 Share Posted October 26, 2014 At a quick glance, you should be balancing this using half equations rather than how you seem to be doing it. Are you familiar with how to do this at all? Link to comment Share on other sites More sharing options...
Sensei Posted October 26, 2014 Share Posted October 26, 2014 (edited) You need as many HCl as NaNO3 on input Because one of products will be NaCl Edited October 26, 2014 by Sensei Link to comment Share on other sites More sharing options...
rainingspiders Posted October 26, 2014 Author Share Posted October 26, 2014 (edited) @ Empress: Yeah, I have tried it and it didn't work out for me. However, I just finished solving it, using the algebraic method. Thanks for the help though! Should I post how I solve it here so other people can see? Edited October 26, 2014 by rainingspiders Link to comment Share on other sites More sharing options...
Robittybob1 Posted October 26, 2014 Share Posted October 26, 2014 @ Empress: Yeah, I have tried it and it didn't work out for me. However, I just finished solving it, using the algebraic method. Thanks for the help though! Should I post how I solve it here so other people can see? yes please Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted October 26, 2014 Share Posted October 26, 2014 @ Empress: Yeah, I have tried it and it didn't work out for me. However, I just finished solving it, using the algebraic method. Thanks for the help though! Should I post how I solve it here so other people can see? Sure. I'm curious to know where you ran into trouble with the half equation method? Did you remember to exclude spectator ions? Link to comment Share on other sites More sharing options...
rainingspiders Posted October 26, 2014 Author Share Posted October 26, 2014 Ok, I'll try my best. This is my first time trying to explain Chem so sorry if it's a bit confusing. First, I assigned a different variable for each compound of the equation. The numbers of these variables will be the coefficients of their corresponding compounds. So... HCl = a As2O3 = b NaNO3 = c H2O = d NO = e H3AsO4 = f NaCl = g I then found out how much of each element is on the reactant side and how much of each element is on the product side, with respect to the variables. Using the following equation... # element X on reactant side = # element X on product side For H, we have "a" amount of H and "2d" amount of H, so we get a + 2d amount of h on the left side. We have 2d because for every compound d, we have 2 H's. Do the same on the product side, and repeat for all the elements you have. Doing so gets us... H: a + 2d = 3f Cl: a = g As: 2b = f O: 3b + 3c + d = e + 4f Na: c = g N: c = e I set a variable equal to 1. I chose the variable "a", just because. Doing so gives us a = 1 g = 1 c = 1 e = 1 so... H: (1) + 2d = 3f Cl: (1) = (1) As: 2b = f O: 3b + 3(1) + d = (1) + 4f Na: (1) = (1) N: (1) = (1) The only functions we have that have variables in them are... Equation 1: 1 + 2d = 3f Equation 2: 2b = f Equation 3: 3b + 3 + d = 1 + 4f , which simplifies to 3b + d = 4f - 2 We want to solve the system of equations. We know what f is, so substitute 2b for f into equation 1 and 2. Equation1: 1 + 2d = 3(2b) 1 + 2d = 6b Equation 3: 3b + d = 4(2b) -2 3b + d = 8b - 2 d + 2 = 5b Solve the system of equations for d. 1 + 2d = 6b >>> -5 - 10d = -30b d + 2 = 5b >>>> 12 + 6d = 30 b combines to 7 + 4d = 0 d = 4 plug in d into equation 1, solve for b, then use be to solve for f. Doing so gives us d = 7/4 b = 3/4 f = 3/2 So listing all our coefficients... a = 1 b = 3/4 c = 1 d = 7/4 e = 1 f = 3/2 g = 1 Multiply these values by the least common denominator, which is 4. This gives us... a = 4 b = 3 c = 4 d = 7 e = 4 f = 6 g = 4 Write out the chemical equation with the corresponding coefficients. [latex] 4HCl + 3As2O3 + 4NaNO3 + 7H2O -> 4NO + 6H3AsO4 + 4NaCl [/latex] If we check, we see that the equation balances. Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted October 26, 2014 Share Posted October 26, 2014 Here's what I get with half equations. Excluding spectator ions, the equation comes to: H+ + As2O3 + NO3- + H2O --> NO + H3AsO4 The As goes from As (3+) to As (+5) and the N from N (+5) to N (+2) Oxidation: As2O3 + 5H2O --> 2H3AsO4 + 4H+ + 4e Reduction: NO3- + 4H+ + 3e --> NO + 2H2O Balancing the electrons gives: 3As2O3 + 15H2O --> 6H3AsO4 + 12H+ + 12e 4NO3- + 16H+ + 12e --> 4NO + 8H2O Combining: 3As2O3 + 7H2O + 4NO3- + 4H+ --> 6H3AsO4 + 4NO Adding the spectator ions back: 3As2O3 + 7H2O + 4NaNO3 + 4HCl --> 6H3AsO4 + 4NO + 4NaCl Edit: Your method seems to have worked (I haven't checked it), but it seems awfully complicated. Where were you going wrong with the half equations? Link to comment Share on other sites More sharing options...
rainingspiders Posted October 29, 2014 Author Share Posted October 29, 2014 Hey, sorry for the late reply. I had midterms this week so I wasn't able to go on the internet that often. Anyways, it turned out I didn't take the spectator ions into account. Link to comment Share on other sites More sharing options...
chemistry student Posted October 30, 2014 Share Posted October 30, 2014 Can't believe you guys do all of this I just do it by trial and error. What stage of education are you at? I'm doing my A-levels :3 Thanks though! I've found my way of balancing this really hard question using the algebra technique :'D Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted October 30, 2014 Share Posted October 30, 2014 Trial and error is a very poor way if approaching a question like this. If you use the appropriate methods, you will spend much less time on it. Link to comment Share on other sites More sharing options...
chemistry student Posted October 30, 2014 Share Posted October 30, 2014 Trial and error is a very poor way if approaching a question like this. If you use the appropriate methods, you will spend much less time on it. Ye but what I meant was there's this many x atoms there so same on the other sides. There's this number of y atoms here so let there be y atoms of this on the other side. Then I erase it and start again. And ye I guess it depends on the question. Trial and error is a much quicker method Link to comment Share on other sites More sharing options...
Sensei Posted November 22, 2014 Share Posted November 22, 2014 [latex] 4HCl + 3As2O3 + 4NaNO3 + 7H2O -> 4NO + 6H3AsO4 + 4NaCl [/latex] If we check, we see that the equation balances. Correct. My new application for chemistry is also showing it's good answer, see: Link to comment Share on other sites More sharing options...
Fuzzwood Posted November 22, 2014 Share Posted November 22, 2014 That's because you don't start out from the half equations. If you can figure out those, it's usually just balancing electrons after that. The rest falls into place. Link to comment Share on other sites More sharing options...
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