H2S+KMnO4+H2SO4---K2SO4+S8+MnSO4+H2O

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- Nov 15th 2006, 03:31 PMjerryramosCan anyone help me balance this? CHEM
H2S+KMnO4+H2SO4---K2SO4+S8+MnSO4+H2O

- Nov 15th 2006, 04:19 PMtopsquark
If you can't think of any better way to balance one of these beasts you can always do this:

$\displaystyle aH_2S + bKMnO_4 + cH_2SO_4 \to dK_2SO_4 + eS_8 + fMnSO_4 + gH_2O$

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:

$\displaystyle 36H_2S + 48KMnO_4 + 44H_2SO_4 \to 24K_2SO_4 + S_8 + 48MnSO_4 + 80H_2O$

-Dan