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    Iron is extracted by heating iron oxide with carbon monoxide.

    Fe2 O3 + 3CO -----> 2Fe + 3CO2

    a) Calculate the theoretical yield of iron that can be obtained when 320kg of iron oxide (Fe2 O3) reacts with excess carbon monoxide.

    b) The actual yield of iron is 100kg. Calculate the percentage yield of iron.

    This is really hard and not straight forward, I have got an answer that I have worked out, but not sure if it's right, could a really good chemist student work it out and see what you got? My answers are in the spoilers. Thanks!!

    Spoiler:
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    a) 146.9kg
    b) 68.1%
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    Sorry you've not had any responses about this. Are you sure you've posted in the right place? Here's a link to our subject forum which should help get you more responses if you post there.
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    I got a different set of answers to you. This was my method:moles= mass / molecular mass so the number of moles of Fe2O3 is:320000g / 159.6 gmol-1 = 2005 molThen the ratio is 1:2 between the Fe2O3 and the Fe so you get 4010 mol of Fe. To get the mass you do 4010 x 55.8 = 223758g = 223.758kgWith this answer the percentage yield is 44.7%.
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    (Original post by hannah-rosie)
    I got a different set of answers to you. This was my method:moles= mass / molecular mass so the number of moles of Fe2O3 is:320000g / 159.6 gmol-1 = 2005 molThen the ratio is 1:2 between the Fe2O3 and the Fe so you get 4010 mol of Fe. To get the mass you do 4010 x 55.8 = 223758g = 223.758kgWith this answer the percentage yield is 44.7%.
    i got the same answer. would it make a difference if u didn't convert from g to kg?
 
 
 
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