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# Sodium ferrate Watch

1. 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

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).
Express your answer to an appropriate number of significant figures.
2. (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

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).
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 = ?
3. (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

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).
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%

4. (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.
5. (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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Updated: October 15, 2017
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