Alain12345
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Chemistry Homework Help
Can someone help me with this question please?
a) How many grams of FeS are necesary to react with 7.81 g of oxygen according to the following equation?
4FeS + 7 O2---> 2Fe2O3 + 4SO2
b) How many moles of O2 are necessary to react with 6.79 mol of FeS?
I tried it, but I'm not sure if I got it right. My answers are:
a) 12.3 g
b) 1.22x10^3 mol
[Edited on 23-11-2005 by Alain12345]
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Magpie
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b) Are you saying its going to take over a 1000 moles of O2 to react with 6.79 moles of FeS? Does this sound reasonable?
Hint: Each specie in the equation can be treated as a mole. Each mole can be considerd to weigh its molecular weight.
The single most important condition for a successful synthesis is good mixing - Nicodem
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The_Davster
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First answer is correct(and it is slightly harder compared to the second), the second is however incorrect, does over 1000moles of O2 reacting with
under 10 moles of O2 seem correct to you?
EDIT: Magpie just beat me to b)  .
[Edited on 23-11-2005 by rogue chemist]
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Alain12345
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lol I was pretty sure the second one wasn't right. Thanks a lot.
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BromicAcid
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For the second one here is one way to arrive at the answer. Each reaction takes 4 mol of FeS, you have 6.79 mol, divide this by 4 and you get 1.6975,
that is how many times the reaction will run, each time it is run it consumes 7 mols of oxygen. Multiply the number of times the reaction will run
through, by the amount of oxygen consumed each time to get your answer, in this case 11.88 mol of O<sub>2</sub> are consumed for every
6.79 mol of FeS present.
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Quibbler
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Just a thought how do you weigh 7.8 g of oxygen. Typical chem teacher not a real problem just made up and of little practical use. Oxygen would be
measured by volume if at all.
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Darkblade48
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| Quote: | Originally posted by Quibbler
Just a thought how do you weigh 7.8 g of oxygen. Typical chem teacher not a real problem just made up and of little practical use. Oxygen would be
measured by volume if at all. |
Chances are they haven't learned the (ideal) gas law yet, since they're still doing stoichiometry 
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Thomas Winwood
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| Quote: | Originally posted by Darkblade48
| Quote: | Originally posted by Quibbler
Just a thought how do you weigh 7.8 g of oxygen. Typical chem teacher not a real problem just made up and of little practical use. Oxygen would be
measured by volume if at all. |
Chances are they haven't learned the (ideal) gas law yet, since they're still doing stoichiometry |
Indeed. That said, far be it from me to fail to assist a fellow chemist...
To tackle the problem you need to know that one mole of any gas takes up 22.4 cubic decimetres of volume at 273 K and 1 atm pressure. From that you
can plug in the molar mass of oxygen and work out how much space 7.8g would occupy.
At 273 K, 16g of O<sub>2</sub> takes up 22.4dm<sup>3</sup>, so 7.8g takes up (22.4/16)*7.8 = 10.92 dm<sup>3</sup>.
(Remember that PV = nRT, so the volume one mole of gas takes up changes with temperature.)
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Quibbler
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I think it is done at 298 K these days so 1 mole (ideal gas) occupies 0.0248 m3. There's even some smart-asses who use 1 bar = 100 kPa (not 1 atm
= 101.3 kPa) so we are looking at 0.024465 m3. This has caught me out on many occasions.
My point was why not choose a reaction using liquids/solids. You might put the idea in impressionable young minds that professional chemists go around
measuring out 7.8 g of oxygen.
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Alain12345
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I have another question, so I'll post it here.
Calculate the mass of nitric acid that is produced if 4.00 mol of ammonia is completely reacted through the Ostwald process.
4 NH3 + 5 O2 ---> 4 NO + 6 H2O
2 NO + O2 ---> 2 NO2
3 NO2 + H2O ---> 2 HNO3 + NO
I tried it, but I'm pretty sure I'm way off. How would a solve this question? I thought I could do it by using only the last reaction, but
it didn't work out.
Thanks.
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Darkblade48
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| Quote: | Originally posted by Alain12345
I have another question, so I'll post it here.
Calculate the mass of nitric acid that is produced if 4.00 mol of ammonia is completely reacted through the Ostwald process.
4 NH3 + 5 O2 ---> 4 NO + 6 H2O
2 NO + O2 ---> 2 NO2
3 NO2 + H2O ---> 2 HNO3 + NO
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With this, you simply have to work out each equation.
I'll explain the general concept:
With 4 NH3 + 5 O2 ---> 4 NO + 6 H2O
You start with 4 moles of NH3, so going through the reaction, you will end up with 4 moles of NO and 6 moles of water.
In the second step, you will use your number from above for NO (4 moles) and plug it into this equation:
2 NO + O2 ---> 2 NO2
You will then get the number of moles of NO2.
Finally, you use the number obtained and put it into this equation:
3 NO2 + H2O ---> 2 HNO3 + NO
And solve for the number of moles of HNO3.
Remember that the coefficients in front of the chemicals are merely ratios.
I.e. for 3 NO2 + H2O ---> 2 HNO3 + NO, it means for every 3 moles of NO2 you put in (along with 1 mole water), you will get back out 2 moles of
HNO3 (and 1 mole NO).
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