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[mecramarathon]

The heat of combustion for ethanol is 1367 Kj/mol. Calculate the energy produced per tonne of carbon dixiode from the combustion of ethanol.

The pH of a 0.1 mol/L solution of a monoprotic acid was measured by a student and found to be close to 3. What proportion of the acid molecules remains uncoverted to ions? (
A) 0% B) 1% c) 60% d) 99%

I'll rep u if u can solve it!

 

[adomad]

write the equation
C2H6O+ 3O2 -->3H2O+ 2CO2

in on tone of co2, there is 22727.27273 moles.
divide by 2 to get the moles of ethanol (mole ration from balanced equation) and moles ethanol = 11363.63636.
times this by the delta H given in the equation...


11363.63636 X 1367 =15.53x10^6 kj/ tonne

think that is right

and the second one is D.

 

[ar7]

to find this use establish the number of moles of CO2:

1000kg = 1 000 000 grams

n = 1 000 000/44(molecular weight of CO2)
n = 22727.27 moles of CO2 gas

then find the combustion formula for ethanol:
2C2H5OH + 5O2 --> 4CO2 + 3H2O
2: 5: 4: 3

then as there is double the ratio of moles of CO2 to ethanol produced during complete combustion:

n(ethanol) = 22727.27 /2
n= 11363.64

then multiple the number of moles of ethanol by the molar heat:

1367 * 11363.64 = 15 534 095.88 kJ per tonne of CO2 produced

 

[ar7]

damn u beat me adomad

 

[mecramarathon]

well done both of you! (gosh should have thought of that...)

could u or someone do the second q? plz thanks!

 

[adomad]

well the pH was 3 yeah.... so that means that the [H+]= 10^-3.....(Common rule [H+]=10^-pH)

if the acid was a strong acid for example .1 M HCL. the ionisation will be
HCL--> H+ + Cl- and useing your mole ratios... the concentration of [H+]= .1

do ([hydrogen ions] in the weak acid)/([hydrogen ions] in a strong acid which completly ionises) = 10^-3/.1 X 100 = ionisation percentage


this equals one... hence it follows that the intact molecules are 99% since only one percent ionised

 

[adomad]

damn u beat me adomad

lol... should be doing my english essays but <3 chem lol