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Math Help - Can anyone help me balance this? CHEM

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    Can anyone help me balance this? CHEM

    H2S+KMnO4+H2SO4---K2SO4+S8+MnSO4+H2O
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  2. #2
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    Quote Originally Posted by jerryramos View Post
    H2S+KMnO4+H2SO4---K2SO4+S8+MnSO4+H2O
    If you can't think of any better way to balance one of these beasts you can always do this:
    aH_2S + bKMnO_4 + cH_2SO_4 \to dK_2SO_4 + eS_8 + fMnSO_4 + gH_2O
    where the a, ..., g are your balancing coefficients.

    Now work through each element on each side of the reaction:
    For H:
    2a + 2c = 2g

    For S:
    a + c = d + 8e + f

    etc.

    Now solve the system of equations:
    2a + 2c = 2g
    a + c = d + 8e + f
    b = 2d
    b = f
    4b + 4c = 4d + 4f + g


    f = b, so:
    2a + 2c = 2g
    a + c = d + 8e + b
    b = 2d
    4c = 4d + g

    From the bottom: g = 4c - 4d so:
    2a = 6c - 8d
    a + c = d + 8e + b
    b = 2d

    b = 2d, so:
    2a = 6c - 8d
    a + c = 3d + 8e

    From the bottom, c = 3d + 8e - a so:
    8a = 10d + 48e

    4a = 5d + 24e

    e = (1/24)(4a - 5d) = a/6 - 5d/24

    Now we need to pick some convenient values for a and d:

    I'll try a = 36, d = 24 (These are the smallest values that make e an integer and positive):

    e = 1, c = 44, b = 48, g = 80, f = 48

    So:
    36H_2S + 48KMnO_4 + 44H_2SO_4 \to 24K_2SO_4 + S_8 + 48MnSO_4 + 80H_2O

    -Dan
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