Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Vultux on February 12, 2017, 08:22:19 PM
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Methane is a very powerful greenhouse gas. One pound of methane traps 25 times more heat in the atmosphere than a pound of carbon dioxide. Methane is also the main ingredient in natural gas. Because methane can be captured from landfills, it can be burned to produce electricity, heat buildings, or power garbage trucks. What volume of oxygen is necessary to completely react with 8.46 x 1019 molecules of methane gas, CH4 during combustion?
So I started with balancing the equation, which is CH4 + 2O2 :rarrow: CO2 + 2H2O
I think next I'm supposed to go from 8.46 x 1019 particles of methane gas (CH4) to moles of (CH4) to moles of oxygen to volume of oxygen. What I'm confused about is how I'm supposed to set the equation up and I would appreciate any help in doing so.p
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Start from balanced reaction
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Now change number molecules to moles and calculate moles of oxygen in relation to that of methane.
Hint - Avogadro number.
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So I divide particles by Avagadros which becomes 1.40*10-4 moles. I'm confused on how I do moles to moles. I know it requires the molar mass of both of them but I'm confused on how to do so.
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No, you don't need molar masses.
Do you know how to read the reaction equation (http://www.chembuddy.com/?left=balancing-stoichiometry&right=stoichiometric-calculations)?
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Not really, my teacher is horrible at teaching so I have to learn everything for Chemistry online by myself. Read it and I'm guessing I have to set a ratio of Oxygen to Methane gas which would be 2/4 then do something?
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Now use Avogadro Law nad take molar volume at STP.
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Not really, my teacher is horrible at teaching so I have to learn everything for Chemistry online by myself. Read it and I'm guessing I have to set a ratio of Oxygen to Methane gas which would be 2/4 then do something?
Use the ratio to find number of moles of oxygen needed to burn the methane. Then follow AWK's advice.
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So it would be (1.40*10-4)/2 which makes 0.00007 moles of Oxygen then what? Do I multiply by Avagadros to get volume?
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(1.40*10-4)/2
???
Do I multiply by Avagadros to get volume?
???
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So it would be (1.40*10-4)/2
No. Look at the reaction equation - every mole of methane needs TWO moles of oxygen. If so, 1.40*10-4 moles of methane need how many?
then what? Do I multiply by Avagadros to get volume?
No. Avogadro's number tells you how many molecules are in the mole, not what volume they occupy.
Do you know how to deal with units? Avogadro's number has units of
[tex]\frac{number~of~molecules}{mole}[/tex]
Number of moles has - not surprisingly - units of just [itex]mole[/itex]. When you multiply the two, you get
[tex]mole\times\frac{number~of~molecules}{mole}[/tex]
Mole cancels out and the result is [itex]number~of~molecules[/itex].
Do you know what is a volume occupied by 1 mole of a gas at STP? Or do you know the ideal gas equation? If not, google these things.
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Would it be 2.8*10-4 moles of Oxygen? and then multiply that 22.4? Sorry if I'm being retarded and dumb, It's just that my teacher taught us conversions the first week during school and it's been a while.
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Would it be 2.8*10-4 moles of Oxygen? and then multiply that 22.4?
Yes.
Sorry if I'm being retarded and dumb, It's just that my teacher taught us conversions the first week during school and it's been a while.
I strongly suggest you get back to conversions then, as you will need them all the time.
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So the answer would be 6.3*10-3 liters of Oxygen?
I strongly suggest you get back to conversions then, as you will need them all the time.
Yeah I know, I'll be learning it after I finish this today.
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So the answer would be 6.3*10-3 liters of Oxygen?
Yes. If we are at it - try to convert it to mL.
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Would it be just 6.3 ml of Oxygen?
also could you help me with this part to the question. I'm confused on how to setup the problem with the methane one.
If the reactant present in the least amount were to be quadrupled, how many grams of each product would be made.
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yes 6.3 ml of oxygen.
What reactant is present in the least quantity? What is 4 times that? Re-evaluate limiting reagents for the reaction with this in mind.
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Would that be CH4? Just basing on the coefficients as I'm not sure what it's supposed to be based on.
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If the reactant present in the least amount were to be quadrupled, how many grams of each product would be made.
Question is not clear to me. I am not sure if it is intended to mean "If the reactant present in the least amount were to be quadrupled without changing amount of the other reactant" or "If the reactant present in the least amount were to be quadrupled and the other reactant is in excess". These are two completely different problems.
I guess it is intended to be a limiting reagent type problem, just poorly worded.
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What would be the limiting reagent in this equation? also how would I find out the amount of grams the products would make?
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What would be the limiting reagent in this equation?
What is a limiting reagent doesn't depend on the reaction, it depends on the relative amounts of substances present. Have you tried to google the term and read?
also how would I find out the amount of grams the products would make?
How do you convert moles to mass?
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I understand what the limiting reagent is, just confused on how to find the mole/gram without some starting point.
To go from moles to mass, you multiply the amount of moles by the molar mass.
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Is the starting point not in the question you posted in your OP, and in the answer you came to for the first question? Or is there something else you aren't posting?
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It's a addition to the question. ""If the reactant in the least amount were to be quadrupled, how many grams of each product would be made?"
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It's a addition to the question. ""If the reactant in the least amount were to be quadrupled, how many grams of each product would be made?"
Right, so the assumption then is you use the amounts you used in the prior question.
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But how would I convert them without some sort thing to convert it with
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But how would I convert them without some sort thing to convert it with
I don't understand your question. Convert what to what?
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For limiting reagents, don't I have to convert them to moles to find out which runs out first or something? I have no idea what I'm doing to be honest.
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For limiting reagents, don't I have to convert them to moles to find out which runs out first or something? I have no idea what I'm doing to be honest.
Did you not calculate this in the question before hand?
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I calculated the moles of Oxygen and methane gas, not the products.
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I calculated the moles of Oxygen and methane gas, not the products.
Okay, so how did you know how many moles of oxygen you needed in that question? What concepts did you use to arrive at that conclusion?
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I started with methane gas molecules and converted that to methane moles. I then used that and set up the mole ratio and used that to get to moles of oxygen.
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I started with methane gas molecules and converted that to methane moles. I then used that and set up the mole ratio and used that to get to moles of oxygen.
Okay then, so if we just take that to start with. The mole ratios you used can also be utilised to work out the moles of products. That's why we put coefficients out front of the products as well as the reactants.
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So the CH4 moles are 1.40*10-4 and Oxygen is 2.8*10-4. Do I convert them to grams and find the lowest number?
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So the CH4 moles are 1.40*10-4 and Oxygen is 2.8*10-4. Do I convert them to grams and find the lowest number?
The question is a bit vague on that front. It wouldn't matter, though. It's going to be the same whether you take the lowest number of moles, or lowest number of grams.
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So it's CH4. Now after I multiply it by 4, what do I do next?
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So it's CH4. Now after I multiply it by 4, what do I do next?
I'm not doing your assignment for you. Refer to previous posts in this thread on limiting reagents, as well as the posts in the other forum you are posting in.