Asked by Kaylene

Deterioration of buildings, bridges, and other structures through the rusting of iron costs millions of dollars everyday. Although the actual process also requires water, a simplified equation (with rust shown as iron(III) oxide (Fe2O3) is:

4Fe(s)+3O2(g) --> 2Fe2O3(s)

delta(h reaction =-1.65 x 10^3

How much heat is evolved when 0.250 kg of iron rusts?

How much rust forms when 2.00 x 10^3 kJ of heat is released?

I have tried everything and aris is not accepting it! Please tell me what I'm doing wrong!
Answered by DrBob222
If you showed me what you are doing, perhaps I can figure out what you're doing wrong. Without work, though, I'm in the dark what you're doing wrong.
Answered by DrBob222
It could be as simple as a unit thing. You don't know units on delta H from the problem. That's 1.65 x 10^3 what?
Answered by Kaylene
It's 1.65 x 10^3 kJ.

This is what I did for the 1st question:

-0.250 kg Fe x 1g Fe/10^3 kg x 1 mol Fe/223.4 g Fe x -1.65 x 10^3 kJ/4 mol Fe = 4.61 x 10^-4 kJ Fe


2nd question: 2.00 x 10^3 kJ Fe2O3 x 2 mol Fe2O3/ -1.65 x 10^3 kJ x 319.4 g Fe2O3/1 mol Fe2O3 = -774.3 g Fe2O3
Answered by DrBob222
I got to the first factor and stopped.
0.250 kg is 0.250 x (1000 g/kg) = 250 grams, not 0.250 x 10^-3 grams. 1 g iron is not equivalent to 10^3 kg. The factor is 1 kg/10^3 grams, not 1 g/10^3 kg. Also, haven't you corrected for the 4 moles Fe twice; once with the 1 mol Fe/223.4 and another time with the 1.65 x 10^3/4 mol Fe. I think the answer is about 1846 or so kJ but check it out.
0.250 kg x (1000 g/1 kg)x (1 mol Fe/55.847) x 1.65 x 10^3 kJ/4 mol = ?? kJ.
Answered by Sharisse
If 4.00 kg of Fe2O3 are available to react,how many moles of CO are needed?
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