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Topic: Stoichiometry Problem  (Read 2673 times)

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Offline Tsiih

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Stoichiometry Problem
« on: September 12, 2009, 11:20:26 PM »
I do not understand what I am doing wrong in this problem. No matter how many times I try, I always get the same answer yet the on line site to submit homework answers says it is wrong.

Problem:

Consider the reaction
3Fe + 4H2O -> Fe3O4 + 4 H2

How much iron is needed for the production of 8.1 ng of Fe3O4? Answer in units of atoms.

My attempt:

8.1 ng [1g / (10^-9)ng] = 8.1 x 10^9 g Fe3O4
8.1 x 10^9 g Fe3O4*(1 mol Fe3O4 / 231.541 g Fe3O4)*(3 mol Fe / 1 mol Fe3O4) = 104949015.5 mol Fe
104949015.5 mol Fe (6.022 x 10^23 atoms Fe / 1 mol Fe) = 6.32 x 10^ 31 atoms Fe



Offline matrix_the01

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Re: Stoichiometry Problem
« Reply #1 on: September 12, 2009, 11:46:26 PM »
What I think you may have done wrong...

You converted the amount Fe3O4 to grams incorrectly.
You said:
"8.1 x 10^9 g Fe3O4"
It should be :
8.1 x 10^-9 g Fe3O4
Remember, 1g = 1 x 10^9ng
Therefore to get from ng's to g's you must say ng x 10^-9
Like to get from 10mm to cm's, you divide 10/10=1cm

What I think you may have done wrong then, is used the formulae wrongly.

n=m/M ( Agreed? )
Where M is molecular mass, and m is the mass of the molecule (grams)

You calculated the molecular mass correctly 231.541 g; however, from there you should say:
n=m/M
n Fe3O4 = 8.1 x 10^-9 g Fe3O4/231.541 g Fe3O4
            = 3.498 x 10^-11 mol

Once you have calculated the number of moles of Fe3O4, you can calculate the number of moles of Fe
1mol Fe3O4 : 3mol Fe
n Fe = n Fe3O4 x 3
       = 3.498 x 10^-11 mol x 3
       = 1.049 x 10^-10 mol

Then you can calculate the number of Fe atoms
1.049 x 10^-10 mol x 6.022 x 10^23 atoms =
AND THE ANSWER IS....
6.319 x 10 ^13 atoms of Fe

Hope that helped 8-o, good luck! :D


Offline Tsiih

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Re: Stoichiometry Problem
« Reply #2 on: September 13, 2009, 12:10:56 AM »
Oh wow! Thank you very much! I did not know that about the nanograms! I tried it again with your suggestion and it turned out to be correct. Thank you once again! I'll keep that in mind.  :)

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