Chemistry question - heat of reaction problem...

[tris-]

this is the question "the following equation describes the burning of ethanol"

CH3OH + 3/2 O2 - > CO2 + 2 H2O

standard heat of reaction = (-726) kJ mol-1

what mass of CO2 would be produced by burning enough methanol to produce 1255kJ of heat.

well i have been trying to work this out, i end up with -78g of CO2 which is impossible, right?

so im not even after the complete answer, but how on earth do you begin to work it out?

who ever helps the most can pick some old skool sweets of choice from the photo i posted, and i will mail them out to you.

 

[swords]

Dont forget the - next to the amount of energy means the reaction is exothermic. IE it releases energy in the form of heat (enthalpy)

So i think you are on the right lines you just confuse the minus with a mathematical negative.

 

[tris-]

i am figuring you would make 1.7x the mass of CO2 already there,

because it would take 1.7x the amount of heat to make 1255kJ of heat.

is this what you would think?

 

[Vladamir]

What level of chemistry is this you're doing out of interest?. I'm doing A2 chemistry atm and i can grasp bits of it

 

[tris-]

A level i assume. i am in a foundation year at uni which starts at GCSE > A level > low degree level shit for the year.

 

[tris-]

anyone got some more info then? using what swords said im still not confident that i am correct

cheers swords for helpin though

 

[Reno]

What sword said is correct. So to even help you further your problem said you need to produce 1255 kJ. The reaction produces 726 kJ/mol in heat. ( the minus in front of this number means you need to put in negative energy to get this reaction to happens which means heat is produced) For every methanol burned you get 1 CO2 so 1 mol methanol=1mol CO2 so you produce 1255/726 = 1.73 mol of CO2. 1 mol of CO2 is 44 grams so 1.73*44=76.1 grams of CO2 is formed

 

[Tilda]

Rep for the chemistry wizz!

 

[tris-]

woo fantastic guys. well if u 2 would like some old skool sweets, take a look at the photo and chose 4 individual items each. pm your address and i will post em out

 

[swords]

I'm amazed i remembered...i haven't had to use that since i did my A-Levels

I'm ok for sweets
Though im hunting for Cola cubes and Butterscotch Ice cream atm...where can i get em from?!?!

 

[tris-]

aquarterof.co.uk

 

[DocWolfe]

aye thats a-level I can remember doing it when I did titrations and crap. I can't remember how to do it though

 

[tris-]

titration, i am doing of those sons of bitches now.
well for the last 3 hours, i been trying to doit. makes no sense at all, im now in talks with a really clever maths dude to try work it out lol.

 

[tris-]

btw if u r feeling bored hereis the titration question

an oven cleaner contained NaOH. the cleaner was analyed by titration with HCL acid containing 84 g dm-3 of hydgoren chloride. 10cm3 of HCL acid were required to neutralise 40cm3 of oven cleaner the reaction equation is

NaOH + HCL ---> NaCl + H2O

calculate the concetration of HCL acid in mol dm-3
calculate moles of HCL acid reacting
calculate concentration of NaOH in mol dm-3
calculate mass of NaOH

 

[Dandare]

btw if u r feeling bored hereis the titration question

an oven cleaner contained NaOH. the cleaner was analyed by titration with HCL acid containing 84 g dm-3 of hydgoren chloride. 10cm3 of HCL acid were required to neutralise 40cm3 of oven cleaner the reaction equation is

NaOH + HCL ---> NaCl + H2O

calculate the concetration of HCL acid in mol dm-3
calculate moles of HCL acid reacting
calculate concentration of NaOH in mol dm-3
calculate mass of NaOH

I'm way out of my depth here as I ain't got a degree and now know why those other guys are on 10k more than me at work
But is question 2 a trick?
Logically thinking and if this is course work probably not a question to ask.....but how can you calculate the moles of HCL reacting when it is constantly being titrated by the caustic?

 

[tris-]

should be the moles of hcl that did react to neutralise it i think, the doc must of made a mistake in the question

 

[Amildin]

This is AS level stuff, if i wasnt hungover still id have a go. Just done the module with moles etc, only way to get the hang of it is to do them over and over again.

The rest can be worked out by common sense usually if you remember the 3 moles calculations:

Moles = Volume x Concentration For liquids (aq)
1000 (assuming volume is in cm)

Moles = Mass For Solids
Relative Formula Mass

Moles = Volume For Gas
24

Surprised I still remember that

 

[tris-]

if anyone wants a crack at the 2nd question be my guest.

i been doing it for 4 hours with a master of engineering and a chemical engineer and we keep getting fuckin crazy numbers like 56 mol dm3 for the concentration of NaOH. what the hell kind of oven cleaner is that. and bloudy 230 mol dm3 for HCl, sure anyone doing this experiment will get an acid of -2.5 pH. GOD DAMN IT!!

anyone know what we r doing wrong?

 

[swords]

Will do it in a bit, xping atm.

 

[Ormorof]

btw if u r feeling bored hereis the titration question

an oven cleaner contained NaOH. the cleaner was analyed by titration with HCL acid containing 84 g dm-3 of hydgoren chloride. 10cm3 of HCL acid were required to neutralise 40cm3 of oven cleaner the reaction equation is

NaOH + HCL ---> NaCl + H2O

calculate the concetration of HCL acid in mol dm-3
calculate moles of HCL acid reacting
calculate concentration of NaOH in mol dm-3
calculate mass of NaOH

well moles = mass/Mr (or RMM depending on how you were taught)

so 84/36.46094 = 2.304 moles of HCl

concentration = moles/volume *1000

so thats 2.304/10 * 1000 = 230moldm-3 mmmmm obviously very conc acid?

so its a 1:1 reaction so 1 mol of HCl = 1 mol NaOH, so that lets you do

2.304/40 * 1000 = 57.6mol dm-3

the mass would then be moles*RMM giving you 2.304*39.9971 = 92.15g

that does seem rather high, but thats probably why you should wear gloves when using oven cleaner?

thats how i would work it anyway, btw where you doing your foundation degree? and are you looking to move onto a BSc or an MChem next year?

 

[tris-]

i am at university of teesside.

next year i goto on BSc(w/ honours)

just curious, why the interest?

so orm, you got the same answers as us.

but they seem rediculous dont you think? wtf kind of acid are the peoples using to do this titration?! the pH is -2.5!!!

i gues in future i wont think about what is really doable. just go by what the question says and imagine some guy using the strongest acid youve ever seen to fucking neutralise some oven cleaner.

 

[Haggus]

Why would the mass be moles * RMM ?

 

[tris-]

btw if anyone wants an anarlagy for moles -

lets say we got a packet of polos. this packet has 16 mints in it.
we want 32 mints, so we would use 2 packets.

replace packet with mol, replace 16 with another rmm and replace polo/mints with another element etc.

or if u r doing diatomic atoms and you only want half a mol, then you only need half the packet in the equation.

(btw, made this up for the engineer cos he didnt understand a mol. i thought it was good )

 

[swords]

This may well be wrong, it's been a while since i've done this.


Moles = Mass/mW

so 84g solid HCL would contain 84/36.5 = 2.3M of HCL.
so you have a 2.3 Molar solution of HCL.

Dm-3 = L so you have a 2.3 Mol/Dm-3 concentration.

Moles of HCL acid reacting is simply:

10ml = 0.01L
so 2.3 * 0.01 = 0.023M HCL in 10ml

If it takes 10ml of this solution to neutralise 40ml of the Caustic solution, the ratio of HCL to NaOH in moles is 1:4

so if you have a 2.3Molarity HCL solution the Molarity of your NaOH is going to be 2.3/4 = 0.575Molarity (Mol/Dm-3) NaOH solution

to get a 1M solution of NaOH you would need to add 40g anhydrous NaOH to 1L H20 (23+ 16 +1) so simply times 40g by the Molarity ( 0.575 * 40 = 23g)

Mass of NaOH in 1L of the Cleaning Solution is 23g

 

[Ormorof]

Why would the mass be moles * RMM ?

it just is

if i could draw the calculation "pyramid" i would, but cba

and tris im just curious, as im doing an MChem at uni of Hull

edit: just read swords answer though, makes more sense

edit again: and its probably correct as ive just noticed that the question asks for the amount of moles that react rather than the actual moles

 

[swords]

I hate DM-. g/L is so much easier.
Then again i am a Biochemist rather than a pure chemist.

 

[tris-]

swords numbers seem more realistic.

but the way orm and i have done it is the formula like c = n/v to get c.

i dont think its as simple to say c is just the moles and then how much liquid is there.

as the volume increase, then theoretically the concentration should decrease, is that right?

so if we go back to the packet of polos, this polo packet contains 36.5 polos.

you put it into a glass of water 10cm3 in volume.

now you take that polo packet and put it in a bucket of water that is 1 litre of volume.

its now harder to find the polos because there is more liquid there.

which is how i see concentration anyway.

or using cordial juice, you ad 5ml to a shot glass and fill the rest with water. its of a high concentration.

you put the 5ml into a litre glass of water now. it tastes weaker.

 

[swords]

I also used that equation in a way.

I assumed the volume of the starting HCL solution to be 1L.
Then i thought about adding 84g of solid HCL to that 1L.
what would the molarity of that 1L be if i added that much HCL mass?
easy, just work out the number of moles in 84g solid HCL.

which you know from moles = mass /mW

that gives you an easy base to work everything else out.

 

[tris-]

thats all well an good man, but the volume is 10cm3 only. nothing else.

 

[swords]

Moles of HCL acid reacting is simply:

10ml = 0.01L
so 2.3 * 0.01 = 0.023M HCL in 10ml

multiply the Molarity by the volume and you get the number of moles in that volume solution.
You can apply this to a 1L solution, 2L or 10ml if you like.