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    2H2S + SO2 -> 3S + 2H20

    △H: -145.6 kJmol^-1

    △fH:

    H2S = -20.6
    S02 = ?
    S = 0
    H20 = -241.8

    This is what I did:

    First I rearranged the equation to give me S02 = (2x-241.8) - (2x-20.6) + 145.6.
    Simplified, this gave me -483.6 - 104.4 to give me -588.

    The actual answer is -296.8 and I'm not sure how to get that so any help would be appreciated, thanks.
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    -145.6 = - (2 x -20.6) - x + (3 x 0) + (2 x -241.8)
    x = -296.8
 
 
 
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Updated: November 7, 2017
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“Yanny” or “Laurel”

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

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