[H^j]

Hey everyone.  Quick question that has me stumped.  I get really close to the answer, but I'm off by a couple numbers big enough to make a difference.  Here it is:

For the balanced reaction for forming iron from ome of its ores, hermatite:

Fe2O3 (s) + 3CO----> 2Fe(s) + 3CO2(g )

calculate the percentage yield if 1150 kg of hermatite is reduced by 706 kg of carbon monoxide to give 635 kg of iron. Which reactant is limiting?

I absolutely cannot figure this out!!!  Any help would be appreciated.

Thanks!

[Alpha-Omega]

HINT:  Species react mole to mole NOT gram to gram...do the conversions...

[IITian]

HINT (well...almost the answer!)

Molecular weight of the species involved:

Fe₂O₃ = 159.69

CO = 28.01

Fe = 55.847

CO₂ = 44.01

your balanced rxn shows that 1 mole of Fe₂O₃ reacts with 3 moles of CO to form 2 moles of Fe and 3 moles of CO₂. I guess you can figue it out...

[Lilly]

Theoratical / Actual yield
Step 1: Calculate the number of moles of Fe2O3.
Mass/Mr=Moles
1150/159=7.23moles
Maximum amount of Fe2O3 that could be made is 7232.7
Step 2: Calculate the number of moles of Iron produced.
Again Mass/Mr=moles
635/55.9=11.36
Step3: Calculate the percentage yield
7.23/11.36 x 100 = 63%
I think i cracked it..I got practical exam based on this tommorow.

[Borek]

1150/159=7.23moles

OK

Maximum amount of Fe2O3 that could be made is 7232.7

Doesn't make sense. First, you are not producing Fe₂O₃, second, no idea what is this number.

Step 2: Calculate the number of moles of Iron produced.
Again Mass/Mr=moles
635/55.9=11.36

OK

Step3: Calculate the percentage yield
7.23/11.36 x 100 = 63%

No. First - percentage yield is actual to theoretical, not theoretical to actual, second - theoretical is not 7.23 mole.

[Lilly]

No. First - percentage yield is actual to theoretical, not theoretical to actual, second - theoretical is not 7.23 mole.

I don't know this is the procedure i read in the book. It was actual yield/maximum yield x 100.

How do you suggest it should be done? ..If i am wrong this will help me out.

[Borek]

Both procedures - mine and from the book - call for actual yield in nominator. You have put actual yield in denominator. Theoretical and maximal yields are the same in this case, they differ just by name.

[Lilly]

Both procedures - mine and from the book - call for actual yield in nominator. You have put actual yield in denominator. Theoretical and maximal yields are the same in this case, they differ just by name.

OMG such a silly mistake, if i had done that in the practical exam.