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!
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.
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?
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
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.
If 4.00 kg of Fe2O3 are available to react,how many moles of CO are needed?
To solve these problems, you need to use stoichiometry and the given information.
1. How much heat is evolved when 0.250 kg of iron rusts?
To determine the heat evolved, you can use the molar heat of reaction (delta H reaction) and the molar mass of iron. The molar mass of iron is 55.845 g/mol.
First, convert the given mass of iron to moles:
0.250 kg = 250 g
250 g / 55.845 g/mol = 4.47 mol
According to the balanced equation, the molar ratio between iron (Fe) and the heat evolved (delta H) is 4:1.
Therefore, the heat evolved when 0.250 kg of iron rusts is:
4.47 mol Fe x (-1.65 x 10^3 kJ/mol) = -7.37 x 10^3 kJ (note the negative sign, indicating heat released)
2. How much rust forms when 2.00 x 10^3 kJ of heat is released?
To determine the amount of rust formed, you need to use the molar mass of iron(III) oxide (Fe2O3).
First, calculate the moles of heat released:
2.00 x 10^3 kJ / (-1.65 x 10^3 kJ/mol) = -1.21 mol (again, note the negative sign)
According to the balanced equation, the molar ratio between iron(III) oxide (Fe2O3) and iron (Fe) is 2:4. This means that for every 2 moles of Fe2O3, 4 moles of Fe are consumed and vice versa.
Therefore, the moles of iron(III) oxide (rust) formed can be calculated by multiplying the moles of heat released (-1.21 mol) by the molar ratio:
-1.21 mol x (2 mol Fe2O3 / 4 mol Fe) = -0.605 mol (again, note the negative sign)
Finally, convert the moles of rust to grams using the molar mass of Fe2O3 (159.6882 g/mol):
-0.605 mol x 159.6882 g/mol = -96.51 g (again, note the negative sign)
So, when 2.00 x 10^3 kJ of heat is released, approximately 96.51 grams of rust is formed.
Make sure to double-check your calculations and consider the proper use of units throughout the problem. Good luck!