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    I've seen a couple of threads with a similar question but no one has answered it. I need some help making a balanced equation for the reaction :confused:

    14. An unusual compound of iron has been detected on the surface of the planet Mars. This compound contains the ferrate(VI) ion.
    A student uses 1.00 g of iron(III) oxide and makes, on crystallisation, 0.450 g of sodium ferrate(VI), Na2FeO4.
    Calculate the percentage yield, by mass, of sodium ferrate(VI).
    Show your working.
    Express your answer to an appropriate number of significant figures.
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    (Original post by Ry_p94)
    I've seen a couple of threads with a similar question but no one has answered it. I need some help making a balanced equation for the reaction :confused:

    14. An unusual compound of iron has been detected on the surface of the planet Mars. This compound contains the ferrate(VI) ion.
    A student uses 1.00 g of iron(III) oxide and makes, on crystallisation, 0.450 g of sodium ferrate(VI), Na2FeO4.
    Calculate the percentage yield, by mass, of sodium ferrate(VI).
    Show your working.
    Express your answer to an appropriate number of significant figures.
    without other by-products, you see that ratio of Fe2O3 : Na2FeO4 would be 1:2 (i.e 2 Fe atoms on iron (III) oxide reactant would give balanced eqn with 2 Fe atoms on the product side)

    mol of Fe2O3 reacted = 1.00/RMM of Fe2O3

    mol of Na2FeO4 = 2/1 *mol of Fe2O3 = mass of Na2FeO4 produced/RMM of Na2FeO4

    Then % yield of Na2FeO4 = ?
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    (Original post by Ry_p94)
    I've seen a couple of threads with a similar question but no one has answered it. I need some help making a balanced equation for the reaction :confused:

    14. An unusual compound of iron has been detected on the surface of the planet Mars. This compound contains the ferrate(VI) ion.
    A student uses 1.00 g of iron(III) oxide and makes, on crystallisation, 0.450 g of sodium ferrate(VI), Na2FeO4.
    Calculate the percentage yield, by mass, of sodium ferrate(VI).
    Show your working.
    Express your answer to an appropriate number of significant figures.


    The balanced equation is
    Fe203 + 2NaOH ->>> 2Na2FeO4 + H20

    The first step will be to work out the mr of both the compounds:
    Na2Fe04 = 165.8
    Fe203 = 159.6

    Next find mol of fe203 that reacted:
    1.00/159.6
    = 0.00627 mol

    Then use stoichiometry:
    1:2 ratio

    Hence times the moles to get moles of Na2FeO3:
    0.00627 x 2
    = 0.01254mol

    Find the mass of Na2FeO4:
    0.01254 x 165.8
    = 2.07g

    Then use equation for percentage yield:
    (0.450/2.07g)x100
    = 21.7%

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    (Original post by Chelsebry)
    The balanced equation is
    It is lovely of you to solve someone's problem, but I would suggest not (probably) wasting your time solving a four year old problem. The original poser of the Q will have finished whatever course they were on, I doubt they'll care too much about this bit of homework.

    On the other hand, I would also suggest not just solving someone's problem, it is much better to help them solve the problem for themselves.
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    (Original post by Pigster)
    It is lovely of you to solve someone's problem, but I would suggest not (probably) wasting your time solving a four year old problem. The original poser of the Q will have finished whatever course they were on, I doubt they'll care too much about this bit of homework.

    On the other hand, I would also suggest not just solving someone's problem, it is much better to help them solve the problem for themselves.
    Particularly as the "help" starts with:

    The balanced equation is
    Fe203 + 2NaOH ->>> 2Na2FeO4 + H20
 
 
 
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