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Help with balancing a chemical equation algebraically

  1. Dec 27, 2012 #1
    1. The problem statement, all variables and given/known data

    (a)H2SO4 + (b)NaHCO3 -> (x)Na2SO4 + (y)CO2 + (z)H2O

    I gave each element it`s own equation:
    (1) For hydrogen – 2a + b = 2z
    (2) For sulphur – a = x
    (3) For oxygen – 4a + 3b = 4x + 2y + z
    (4) For sodium – b = 2x

    I then assigned a number to letter 'a'. My book says to begin with the number 12 because it is the most conventionally used.

    I began with equation (2) since it had the least amount of variables and proceeded in this fashion.

    a = 12

    (2)
    a = x
    x = 12

    (4)
    b = 2x
    b = 2(12)
    b = 24

    (1)
    2a + 2b = 2z
    2(12) + 2(24) = 2z
    24 + 48 = 2z
    72 = 2z
    z = 36

    (3)
    4a + 3b = 4x + 2y + z
    4(12) + 3(24) = 4(12) + 2y + 36
    48 + 72 = 48 + 2y + 36
    120 = 84 + 2y
    36 = 2y
    y = 18


    I then divided each of the values by a common divider, in this case '6' in order to get the lowest possible coefficient. I ended up with:
    a = 2
    b = 4
    x = 2
    y = 3
    z = 6

    When I add these to my original formula, it is not balanced:
    (2)H2SO4 + (4)NaHCO3 -> (2)Na2SO4 + (3)CO2 + (6)H2O

    I also tried dividing by 12, 4 and 2 and I am still left with an unbalanced formula.

    The book says that when this happens, I must instead assign the value of 1 to 'a', I did this and got the following results:

    a = 1

    (2)
    a = x
    x = 1

    (4)
    b = 2x
    b = 2(1)
    b = 2

    (1)
    2a + 2b = 2z
    2(1) + 2(2) = 2z
    2 + 4 = 2z
    6 = 2z
    z = 3

    (3)
    4a + 3b = 4x + 2y + z
    4(1) + 3(2) = 4(1) + 2y + 3
    4 + 6 = 4 + 2y + 3
    10 = 7 + 2y
    3 = 2y
    y = 1.5



    As you can see, 'y' is left at 1.5 which is not a whole number and therefore unacceptable. Can someone please tell me what I am missing?
     
  2. jcsd
  3. Dec 27, 2012 #2

    jedishrfu

    Staff: Mentor

    So if you got y as 1.5 then why don't you go back and double your other numbers to keep the balance and have integral values...
     
  4. Dec 27, 2012 #3
    If I double those numbers, I will end up with the original a = 2, b = 4, x = 2, y = 3, z = 6 which would keep my formula unbalanced.
     
  5. Dec 27, 2012 #4

    jedishrfu

    Staff: Mentor

    try picking a=30 ie 2*3*5 in other words w're adding in a new factor

    Actually it seems a=2 would work too which means that the 12 should have worked fine.

    Your mistake is in the 2a + 2b = 2z equation where it should have been a b instead of 2b

    In the words of Shakespeare 2b or not 2b that is the question.
     
    Last edited: Dec 27, 2012
  6. Dec 28, 2012 #5
    I'm curious: what course is this for?
     
  7. Dec 28, 2012 #6

    jedishrfu

    Staff: Mentor

    Looks like HS Chemistry
     
  8. Dec 28, 2012 #7
    I have 2a + b = 2z in that equation. 2a is referring to 2 hydrogen atoms in the first compound of the reaction which I have labeled 'a' and 'b' (literally '1b') is for the single hydrogen atom in the second compound of the reaction. 2z is of course the 2 hydrogen atoms in the last compound of the product.
     
  9. Dec 28, 2012 #8

    jedishrfu

    Staff: Mentor

    Here's your work from the original post:

    Code (Text):


    (1)
    2a + 2b = 2z
    2(12) + 2(24) = 2z
    24 + 48 = 2z
    72 = 2z
    z = 36

     
    It should be 2a+b=2z right? and hence z=24
     
  10. Dec 28, 2012 #9
    Ah, yes. This headache was the result of a typo..
     
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