Suppose I
burn 15 grams of butane in 500 grams of
oxygen gas. One of the reactants will be
completely used up and one will be left over.
1) Which reactant will be used up?
2) Which reactant will be left over?
3) How many grams of this reactant will be left over?
(Hint: Write a balance chemical equation first! Then think it through from there.)
So I looked up butane on the internet and found its formula which is C4H10.
ReplyDeleteThis type of reaction is combustion.
The balanced equation I got was:
2 C4H10 + 12 O2 --> 10 H2O + 8 CO2
((I don't know how to do subscripts on this >_<))
Actually, I think the balanced formula is:
Delete2 C4H10 + 13 O2 ----> 10 H2O + 8 CO2
In your formula the number of oxygen atoms on the reactant side is 24, but on the product side, it is 26. So you just need to change the coefficient in front of the oxygen to 13.
Yeah, you're right. :)
DeleteNice catch, Brant.
DeleteI don't know why but I also made the same mistake as Dana.
DeleteNice explanation Brant.
Since the butane is being burned it is the reactant that will be used up. This means that the reactant that will be left over must be oxygen.
ReplyDeleteIf the reactant that will be left over is oxygen, then doesn't that mean that there will be 500 grams of oxygen will be left over? Or would that mean that since the 15 grams of butane were burned up, then a different mass of Oxygen will have to be left over?
ReplyDeleteYes Dana and Brant, it appears to be a combustion. But wouldn't the balanced equation be more simple and appear as C4H10+9O2------5H2O+4CO2.
ReplyDeleteOH! I see where I went wrong. I had the wrong number of Oxygen. The balanced equation would be 2C4H10+13O2--------10H2O+8CO2
DeleteAccording to the balanced equation, it looks to me as if the reactants are no longer the same. There are no reactants left over. What is left over are the products water, H2O, and Carbon Dioxide, CO2. These products are different from the Butane C4H10 and Oxygen gas O2. That means that no reactant is left over since there are different substances left.
ReplyDeleteBut if we are talking individual things, I guess I can see how there is Oxygen left over. There would be 16 atoms of it and you could find the mass by using avogadro's number and molar mass
DeleteBased on the given information I think the mass of 16 atoms of oxygen would be 4.3e-22. or 4.3x10^-22 grams.
DeleteThis comment has been removed by the author.
DeleteI agree that obviously the products are not the same compounds as the reactants, but according to the Law of Conservation of Mass, don't all the reactants technically have to be left over? Or at least the same amount/type of atoms.
DeleteAll the atoms have to be there, and once you balance the equation they are. But that doesn't mean they are in the same compounds. As reactants the atoms form butane and oxygen, but as products they form water and carbon dioxide.
DeleteSo for reactants left over, is the question asking for individual elements or compounds? I don't quite get what ( type of reactant is being left over). Danny made a good point.
ReplyDeletethe reactants are the H2O and CO2 left over from the combustion reaction. It's the compunds.
DeleteIt's asking for compounds. Which is left over, butane or oxygen? And how much butane or oxygen is left over?
DeleteBy the way, would this be a combustion reaction?
ReplyDeleteyes
DeleteSo to kind of summarize what to do to solve this problem...(please add to this or correct mistakes) (this is just for self clarification)
ReplyDelete1) Write an equation for the word problem
2) Balance the equation
3) Take the ratio of moles of Butane to moles of carbon dioxide
4) Convert to grams
So here's what I did
ReplyDeleteAssuming the butane is the reactant that is all burned up, it must be in the H2O and CO2, right? So I just did what we did in class.
For every 15 g of butane, there is 23 grams of H2O.
For every 15 g of butane, there is 45 grams of CO2.
So that sounds real nice, but I'm not too sure if this is correct. But for now, I'll keep going.
From here, I kind of did some educated guesswork. There's only 15 grams of butane but there is 68 grams of "stuff". Wouldn't the other 53 grams of "stuff" come from the oxygen gas? So would it be safe to assume that 53 grams of O is used meaning 447 grams of O is left? I dunno. That's just my guess.
So basically...
1) Butane will be used up
2) Oxygen will be left over
3) 447 grams of Oxygen
But again, this is a lot of educated guesswork...
A LOT OF EDUCATED GUESSWORK
DeleteMY ARMS ARE OPEN FOR CRITIQUE
Wouldn't we only have to do this for a butane to CO2 ratio? Because isn't butane what the reaction uses and isn't CO2 what the reaction produces?
DeleteBut butane also has Hydrogen molecules. That stuff goes into H2O, so I figured we should do that, too.
DeleteThe H2O and CO2 are only needed to make an equation to know the ratio of moles for butane and oxygen gas. After that they are basically useless I think. We can then find out how many grams of O2 are left after using 15 grams of Butane just from the reactant side of the equation
DeleteBut I got the same about the same answer as you. So I guess I just used a weird method?
DeleteWait Dana how would you get 447 grams of Oxygen? Can you maybe post the equation you used?
DeleteOh snap Dana, that's some pretty educated guesswork! Nice job putting the brain cells together. Weird methods are welcome when based on good thinking! :)
DeleteSince the 15 grams of butane are being burned off from the 500 grams of oxygen, then butane would be the reactant that would be used while oxygen would be the leftover result. I'm a little confused as to how to calculate the grams left over? Could anyone help?
ReplyDeleteSo what you would do is 15gC4H10 x 1molC4H10 / 58.14gC4H10(58.14 is the molar mass of C4H10) x 13molO2 / 2molC4H10(becasue you want to convert to moles of oxygen and using the equation we esablished, there are 13molO2 for every 2molC4H10) x 32gO2 / 1molO2 (because you want to get back to grams). after doing this you are left with 53.66 but you want to use 2 sig figs so your final answer would be 54gO2. then using that you know that you originally had 500gO2 so you subtract them to get the very final answer for part 3 of Mrs. Friedmann's question of 446gO2
DeleteNow you're on the right track.
DeleteI'm confused how about how oxygen and water are involved
ReplyDeleteIt is because it is a Combustion reaction with Hydrogen, carbon and oxygen.
DeleteNo but how do you use them in the actual calculations?
DeleteThe products are only needed to balance the equations, after that I didn't need them at all.
DeleteRight on, Connor.
DeleteSo I got the same combustion equation as you guys did so from there I found that...
ReplyDelete1) Butane will be all used up
2) Oxygen will be left over
3) I have no idea if i am even close but I got 348.8 grams of oxygen by converting from The mass of thing one all the way to the mass of thing 2 by using the coefficients in the balanced problem, but I am not sure if I solved for the right thing.
Ok so I did all the steps correct up until I reached the grams of Oxygen/mole of Oxygen!
Deleteyou may have made a simple math error. One error can screw up everything when it come to chemistry.
DeleteSo true, Melissa. Sigh.
Delete15 g C4H10 x 1 mol C4H10/58.14 g C4H10 x 13 mol O2/2 mol C4H10 x 32 g O2/1 mol O2
ReplyDeleteI used this equation to find out how many grams of oxygen would be used if I had 15 grams of butane. The answer gave me 54 grams of oxygen gas, but that is how much we used so we have to subtract that from 500 to find out how much we have left. I got 446 grams.
a. butane
b. oxygen left over
c. 446 gram of oxygen gas left in the reactant
This is what I ended up with too.
Deleteyea i agree with 446 grams of oxygen gas cuz the balanced equation is 2 C4H10 + 13 O2 ----> 10 H2O + 8 CO2 and take the 15 grams of butane and for 15 grams of butane, you use up 54 grams of oxygen. Therefore the butane gets used up, and 446 oxygen is left in the reactant..... and I think that's what Connor and Dana and Harish and others got...
ReplyDeleteso do we all agree upon 446 grams of oxygen left/??
Since all of us got 446 (I got 447...prob sig fig problem plus i used a weird method) I think it's safe to assume that's the answer.
ReplyDeleteCan someone please take the pleasure of nicely explaining it to everyone else?
I agree that it is probably a sig fig problem because following Connor's equation the exact amount of oxygen gas is 53.6635706914 so when we round to the proper sig figs (considering the lowest amount of significant figures given was one sig fig in the number 500 grams of oxygen gas), shouldn't it technically be 50 g oxygen gas from the equation then 450 g oxygen gas as the final answer?
DeleteIt is a sig fig problem.
DeleteIf you look at my post, the equation i used is on there.
ReplyDeleteyeah, i got that butane will be used up and there will be 446g oxygen left, and at first i got it wrong, but accidentlly flipped the numerator and denominator in one of the parts of the equation Connor used, so thanks connor, helped me fix my mistake.
ReplyDeleteThat's when units help you, b/c you can't flip by mistake when you have to make sure everything cancels the right way.
DeleteI also got that there will be 446 grams of oxygen left after the reaction. I used the same method of using the balanced chemical equation to find the ratio of moles of butane to moles of oxygen. I converted the grams of butane (15) to moles using molar mass and then used the ratio of moles of butane to moles of oxygen to find how many moles of oxygen are present for every 1 mole of butane. I converted those moles of oxygen to grams with the result of 54 grams oxygen gas. The final step I did was subtract 54 from 500 to find out how many grams of oxygen were left over.
ReplyDeleteAfter converting everything using molar mass and ratios like in class I got 500 as my final answer, which doesn't make sense. I did all the same steps?
DeleteOops, forgot to subtract the 54 at the end. Caught my own mistake. Final answer: 446 grams of oxygen after the reaction. The reaction that is used up is butane and that leaves oxygen to be left over.
DeleteSo first I balanced the equation and got 2 C4H10 + 13 O2 ----> 10 H2O + 8 CO2, then I did a ratio of butane used to oxygen used, which is 15:54, meaning butane gets used up, and oxygen is left over. Then i subtracted 54 from 500 and got 446, which I think is what some others got too.
ReplyDeleteI still don't understand why it becomes H2O and CO2--i get the rest of the problem ,is it just because the H and C combine with the O?
ReplyDeleteBrian this chemical reaction is a combustion meaning the products always contain water and Carbon dioxide
DeleteSee Connor's comment above... "The products are only needed to balance the equation..."
DeleteAlso, how do you find the ratio of butane to oxygen used? that is confusing me.
ReplyDelete2 C4H10 + 13 O2 ----> 10 H2O + 8 CO2
Deletethat's the chemical reaction and the coefficiants tell us the ratio of moles to each other. for example, we can tell the ratio of C4H10 to Ox is 2:13.
Couldn't have said it better myself.
DeleteSo if I understand this right,
ReplyDelete13 grams of butane= 2 moles of butane
500 grams of O2= 13 moles of O2
When carbon and hydrogen react with O2, it's a combustion reaction, so the products are CO2 and H2O.
When balanced, it looks like this:
2 C4H10 + 13 O2 ------> 8 CO2 + 10 H20.
Well...you're on the right track, Andrew, but not there yet. 13 grams of butane is not equal to 2 moles of butane but the two numbers are related to each other in the calculation. Can someone clear this up for him?
Delete