# chemistry

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3.) One method used in the eighteen century to generate hydrogen was to pass steam through red-hot steel tubes. The following reaction takes place:
3Fe(s)+4H20(g)--->Fe3O4(s)+4H2(g)

1. What volume of hydrogen at STP can be produced by the reaction of 6.28 g of iron?

2. What mass of iraon will react with 500.0 L of steam at 25 degrees C and 1.00 atm pressure?

3. If 285g of Fe3O4 are formed, what volume of hydrogen, measured at 20 degrees C and 1.06 atm, is produced?

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number 2 is iron*

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First, it's best to make only one problem per post.
All of these are stoichiometry. I'll show you how to do the first one in detail and you can follow that template to do the others. Good luck with your chemistry. It's a lot of fun and it put food on my table for my family and me for 60 years.
One method used in the eighteen century to generate hydrogen was to pass steam through red-hot steel tubes. The following reaction takes place:
3Fe(s)+4H20(g)--->Fe3O4(s)+4H2(g)

1. What volume of hydrogen at STP can be produced by the reaction of 6.28 g of iron?

Step 1. Write the balanced equation. You have that.
3Fe(s)+4H20(g)--->Fe3O4(s)+4H2(g)

Step 2. Convert what you are given to mols.
# mols = g/atomic mass = 6.28/55.8 = 0.112 mols Fe.

Step 3. Convert mols of what you have (in this case Fe) to mols of what you want (in this case hydrogen), using the coefficients in the balanced equation.
0.112 mols Fe x (4 mols H2/3 mols Fe) = 0.150 mols Hydrogen.

Step 4. Convert what you have in step 3 to liters hydrogen.
mols hydrogen x 22.4 L/mol =
0.150 x 22.4 = 3.358 liters which will round to 3.36 liters. I rounded to three significant figures since that is what you have in the 6.28 g Fe in the problem.

Note: You need to go through this problem and rework it because I rounded here and there on the atomic masses and molar masses. You should do them by looking up the atomic masses and molar masses from the periodic table. Second, most often the last step (step 4) asks not for liters but for grams. If that is the case, then just remember that grams = mols (which you will have from step 3) x molar mass. That gives grams instead of liters. If any of the others give you trouble, please repost at the top of the page BUT show your work and tell us in detail exactly what you don't understand.

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Thanks sir bob for the first answer. For the next two it involved atm. Im confused as to which formulas to apply each question looks different. and C is converted to K right...

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right. 1 atmosphere is standard pressure so no extra corrections needs to be made for that. 0 degrees C is standard so a correction must be made for T, In kelvin that is 298 K. Use PV = nRT, plugging in 1 for P, V in liters, R is 0.08205 and T is 298. Solve for n which is number of mols. From there, the problem is worked just as I worked the first one.

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Sorry I didn't post my name. My computer shut down, one reason I'm a little late in answering this, and it took a little time to get things back on track. When I first sign on, I get a name of anonymous unless I type in drbob222.

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