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Third part to a stoichiometry question

  1. Oct 2, 2006 #1
    This is the full question:

    Suppose you roast 2.00kg of Fe3S4. The equation:

    4Fe3S4 + 25 O2 --> 6Fe2O3 + 16SO2.
    a) How many kg of Fe2O3 will be produced?
    b) How many kg of SO2 will be produced?
    c) How many litres of air at 25 degrees C are required for roasting? Assuming air has 23% O2 by mass, and density of 25C is 1.2 g/L.
    The molar masses are Fe3S4 = 295.8 g/mol, Fe2O3 = 159.69 g/mol and SO2 is 64.06 g/mol.

    I have done a and b.
    a) = 1.62kg
    b) = 1.73kg.

    I have NO IDEA how to do C! Please guide!
  2. jcsd
  3. Oct 2, 2006 #2
    I think that since its 23% by mass you can say that O2 is 23g out of a 100g sample of air. with this i got the following equation.

    23gO2( 1L )= 23/1.2= 19.6L
    ( 1.2g )

    Im farely posative about this answer, although i may be incorrect.
  4. Oct 2, 2006 #3
    thank you, i tried it again, and ended up with the same answer. !
  5. Oct 2, 2006 #4
    sry if my equation is not clear it should be 1L/1.2g.
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