A wonderful good evening,
10 g of pentane and how many g of oxygen are needed for complete combustion of the pentane?
I would be very happy to receive an explanation that is as detailed as you can 🙂
(2 votes)
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You must determine the atomic or molar mass of the individual substances and then calculate the mass for the oxygen via a three-set from the 10 g of pentane.
Molar mass Pentan C5H12 (5 C atoms each with 12 yield 60 and 12 H atoms each with 1 together make 72). 8 O2 (16 O atoms each with 16 together yield 256).
10 g by 72 times 256 give 35.5 g of oxygen.
Please look at my invoice above in addition to the question. Where did I make a mistake?
You have slightly rounded up within the whole bill (0.14 is already too much), of the oxygen you only need about 1.11 mol, then 32 times 1.11 the searched 35.5 g.
Oxygen as gas forms molecules from 2 atoms O2, therefore the molar mass 32 and not the individual atomic mass 16 must be used, just 8×O2 (8×32=256).
In my book only in the periodic system of the elements are the rounded values that we always expect.
C5H12 + 8 O2 ⟶ 5 CO2 + 6 H2O
So you need the eight-fold amount of O2. Your m=10 g C5H12 are n=m/M=0.14 mol Pentan, so you need 1.11 mol O2, which are m=nM=35.5 g.
Why can’t you just say that if C5H12 is 10g, the 8-fold amount, so 8O2 is 80g?
Because the molecules don’t weigh as much.
If three people are sitting in each car, you don’t expect them to weigh three times as much as the car.
Got it! Thanks for all your time and your effort!!! 🤩 Wish you a nice evening.
I’m sorry I was counting on my rounded number.
Why do you consider the two at the end of the molar mass? In the case of the molar mass, the lowest number is therefore always observed, right? Why is that?
I have always taken my values from the book, and there is atomic mass in u (rounded) = 16.00. I didn’t hit online.
We have now agreed to n(O2)=1.11 mol. Why do you always have 1.12 mol?
Anyway, that’s just a little thing.
In addition, M(O2)=31.999 g/mol
So m(O2) = n(O2) ⋅ M(O2) = 35.5 g
Perfect, now I understand why n(O2) = 8 * 0.14 mol = 1.12 mol rec.
But why do I count for m(O2) = M * n = 2 * 16,00 g/mol * 1,12 mol?
Why is the 2 of O2 taken into account when calculating the molar mass (M=2*16.00 g/mol)? Can you explain this to me as detailed as this perfect answer?
Yes, if you have a mole of O2 molecules somewhere, you also have two moles of O atoms. But no one has ever asked about this because in your question there are no O atoms.
I don’t think you have a good idea of what “substance” means at all. Finally, it is the number of particles, i.e. the number of molecules, but in multiples of a basic number (6⋅1023). That’s as similar to the umgangssprachige dozen:
C5H12 + 8 O2 ⟶ 5 CO2 + 6 H2O
If you have a dozen (or 19 dozen, or 0.12 dozen) C5H12 molecules, you will need eight dozen (or 152 dozen, or 1.44 dozen) O2 molecules for sales. And that has nothing to do with how many atoms an O2 is composed.
One mole is nothing but a dozen, only with 6⋅1023 instead of 12.
But why is the quantity of O2 not twice as large as the quantity of O? In the clamp is O2, so we also have to calculate the quantity of O2 (in the case of n(O2)=1.12 mol), or not? That’s still something unclear to me. Maybe you got some advice.
Because M(O2) is the molar mass of an O2 molecule, and this is twice as large as the mass of an O atom.
The explanation of why the two are not needed there is very helpful. According to the reaction equation, eight times as much O2 as C5H12 are used. Thanks for the good explanation.
I haven’t understood the second part yet. Why do you have to expect 16.00 g/mol=32 g/mol in the molar mass M=2*? Why is this multiplied by 2 but not above (in the calculation of the quantity of material)? I don’t understand.
If you’re still explaining this to me, I hope I understood the task completely.
Because. Reaction equation eight times as much O2 as C5H12 is used. The Two in O2 is only used when calculating M(O2)≅32 g/mol.
Why don’t you pay attention to the low-level 2 after the O?
So n(O2)=8*2* 0.14 mol=2.24 mol
m(O2) = M * n = 16,00 * 2,24
= 35.84 g
The triangle is just a reminder of YouTube, so to speak you keep your finger on what you want to calculate and then get the equation.
I calculated n(C5H12) = 0.14 mol. And then I need the eight-fold amount of O2. Why can’t I write n(802) = 8 * 0.14 mol = 1.12 (With your number 8 * 0.139 I would have come to 1.11, I just rounded)? Why is the 8 in front of the O2 wrong? I am multiplying there with 8. 🤔
And downstairs, I have falsely multiplied by 8. Without it it would be:
g/mol * 1.12 mol = 35.84 g
But I haven’t understood yet, why
Whether there are C2 molecules or not, is completely irrelevant, they certainly do not occur in the reaction equation.
After you have determined n(C5H12)=0.139 mol, you should write n(O2)=8n(C5H12)=1.11 mol (why is 1.12 mol?) — Your “n(8O2) doesn’t make any senseYou want to know the amount of oxygen O2.
In response, you get m(O2)=n(O2)⋅M(O2) and ready.
I don’t know what you mean by the triangle in which the formula characters m and n⋅M are entered.
I have just added my account to the question. Can you please show me my mistake?
I think you made a mistake or I have a mistake.
My question relates to the calculation of the mass of oxygen.
m[g] = n[mol] * M[g/mol]
Searched: m(8O2)= ?
n(8O2) = 1.12 mol
M(8O2) = 8 * 2 * 16,00 g/mol
= 256 g/mol
m(8O2) = n[mol] * M[g/mol]
= 1.12 mol * 256 g/mol
≅ 286,72 g
Where’s my fault?
I have now calculated:
(10)/(12.01*5 + 1.01*12) ≅0.14 mol.
Count the atoms together. Then comes out for C5H12 M=72.15 g/mol.
I don’t understand the bill yet. The mass m[g] is clear, these are the 10g Pentan. But how do I get to the molar mass M[g/mol] that I have to use in the equation here? I’m still in trouble.
Right. The amount of substance is as much as the number of molecules.
So first I always calculate the amount of substance in mol and then multiply with the calculated factor, in this case with 8?
0.14 mol * 8 = 1.12 mol
And then I’m just calculating the mass in grams, right?
m[g] = M[g/mol] * n[mol]
…
If you want to understand it in general, my video will probably help you:
https://www.youtube.com/watch?v=LmvFeSp7w2k
Thank you very much! Two very small (and really short) questions I have about the video (text you said, see highlighted quotation right down here).
In another question, I was told that I have to count with 1 mol = 22.4 L. There the task was as follows:
“Which quantity of fluorine (in g) takes in a volume of 6 liters under standard conditions?”
Do you have to calculate 1 mol = 22,4 L in this case, since the quantity in g must be calculated in the case of >>normal conditions ≤
What exactly are the standards? Are they always different or are they always identical everywhere? Is there a difference between the term >>normal conditions ≤ and >>ideal gases < <
If you want to see the exact task with my solution, you can get to the question here (my solved task is directly in addition to the question):
It is (complicated) differentiated between standard conditions (1 atm, 0°C), laboratory conditions (1 atm, 20°C), standard conditions (1 bar, 0°C) and others. You have to recreate how it is asked. For standard conditions, with 22.4 l/mol. At school, I always reckon with laboratory conditions and (some rounded) 24 l/mol. Finally, it is only about the pupils who basically understand what we are doing.
Ideal gases have nothing to do with the conditions. The ideal gas is gases which have no interactions between their particles. At school, all gases are considered to be ideal, as the real behavior only deviates minimally.
Clearly, you multiplied the factor 8 twice. The indication “n(8 O2)” is already bullshit. n(O2) = …
It’s about the amount of oxygen, not about 8 oxygen. Take the calculated amount of oxygen (1.12 mol), multiply with the molar mass of O2 (32 g/mol) and finished.
M(O2)=32 g/mol, not 16 g/mol.
In class work, the teacher asked the question of standard conditions and there had to be 22.4 L. Tomorrow is the postwriting class work.
I have come to another solution and I do not know what is right now.
I added the question above with a supplement with the photo of my invoice. Can you give me a feedback where my fault is?
came to 35.5 g for oxygen and I came up to 286,72 g for oxygen (see supplement to the question).