Chemistry
OpenStudy (anonymous):

assuming a 90.0% recovery what mass of iron in grams would actually be produced by 927g Fe2O3?

sam (.sam.):

90.0% of 927g is?

OpenStudy (anonymous):

834.3..?

sam (.sam.):

Is the assumption for recovery is for the mass of iron?

sam (.sam.):

$Fe_2O_3 + 3 CO \rightarrow 2 Fe + 3 CO_2$ Nevermind that, for every mole of $$Fe_2O_3$$ we should get 2 moles of iron, Fe.

sam (.sam.):

So, we should find the moles of iron itself first, by doing $927g Fe_2O_3 \times \frac{1~molFe_2O_3}{159.69~gFe_2O_3} \times \frac{2~mol Fe}{1~molFe_2O_3}=11.61~moles~of~Fe$

sam (.sam.):

Then use the 11.61moles of Fe multiplying it by the molar mass of Fe and you'd get the mass. $11.61molesFe \times \frac{55.85gFe}{1~molFe}=648.4gFe$

sam (.sam.):

Then it says you should be recovering 90% of the iron, so 90% of 648.4 grams of iron is $\frac{90 \%}{100 \%} \times 648.4g=583.6~\text{grams of iron recovered}$

OpenStudy (anonymous):

thanks!

sam (.sam.):

yw :) make sure you're getting it

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