Help PLease- Molar Heat Qu =) (1 Viewer)

[lisarh]

I can't seem to get any of the answers.

ΔH/m x M = 2021

Given m = 0.06 and M = 74.12. Find ΔH
then sub it in the formula?
ΔH= -mcΔt

Am i supposed to do this or not.
I keep getting 21 >.<

Can someone please show me the proper way how to solve it.

 

[lolokay]

from the molar heat of combustion and number of moles of propanoal, you know ΔH.
you know the mass of the water, and the heat capacity of water, so you can rearrange ΔH= -mcΔT to get ΔT

then add that to initial temp

 

[gurmies]

I used the exact same method lolokay, but I can't seem to get the answer. Perhaps you can try it?

 

[Timothy.Siu]

well,

u know

delta H=-mCdeltaT
delta H=-4.18x200xdelta T
thats the heat of combustion

molar heat=(-4.18x200/0.6) x60 x deltaT (i divided by the mass burnt and multiplied by the molar mass)
=83600 delta T joules/mol
=83.6 x delta T kilojoules/mol
2021=83.6xdelta T
delta T=24.2
therefore final temp is 21+24.2=45.2
60 is the mass of propanol

i hope u understand lol, not explained very well

 

[gurmies]

Lol, i used 60.094 as molar mass <>

 

[Lucid Scintilla]

Given m = 0.06 and M = 74.12. Find ΔH

You have one too many zeroes, for starters.

Systematically approach calculatory problems in chemistry. Having standardised your values in SI units, proceed by having the number of moles worked out (via the formula n=m/M).

Then, since you have your molar heat of combustion value given, multiply that by your n ('per mole' will cancel, if you must use this to check if everything's all right).

Having found that, substitute that change in enthalpy in the formula (d)H=|-mc(d)t| (where m is the mass of water).
Solve for t, then add that to your initial temperature value.

I don't want to ruin your fun, kiddo, so you can do the appropriate solving; plus, ... I sort of cannot be bothered.

Spoiler

Fuck it, I'm feeling completionist-perfectionist.

My result is 29.blah, which is obviously wrooong, because the given answers can never be wrong!

I checked through your stuff, and propan-1-ol is 60.10g/mol, not 74.14.
C3-OH (my shorthand for C3H8O; I don't advise using it) is 12.01 x 3 + 16 + 8 x 1.008.

So... with the accurate value of 60.10 as the mass of propan-1-ol/1-propanol, plug your stuff into a calculator, or mentally do it; to achieve the resulting value of 24.13.

Add that to your initial temperature of 21 deg. C, to get 45.2.

Somewhere there I think I did a rounding-too-early error, but whatever.

The 24.2 deg. C answer is there to throw you off, I take it.

I explain terribly because there's no scanner and/or I'm not being paid. >:[

Penes, I sacrifice speed for perfection, and am beaten to the punch by three. x(

 

[lisarh]

Thanks everyone. I carelessly mixed up butanol with propanol. xD

PROPANOL = C3H7OH

n= m/M
n= 0.6/60.094
n= 0.01 moles (2dp)

ΔH= -mcΔt
2021 = (200x4.18)/n . Δt
2021 = (200x4.18)/0.01 . Δt
2021 = 83,730.97 . Δt J/mol
Since 2021 kJ/mol
2021 = 83.73 . Δt
Δt = 24.136

Final temp: 24.136 + 21 = 45.136
Hence (c)

^^