http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2008exams/pdf_doc/2008HSC-chemistry.pdf
Q24 please of the 2008 chem HSC please show working
Q24 please of the 2008 chem HSC please show working
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Chem question help needed, exam tomorrow please help. (1 Viewer)
- Thread starter Twickel
- Start date
shawty-snappin
Member
to find the heat of combustion per gram u gotta divide it by the molar mass
mH (molar heat)/g for petrol = 5460/114 = 47.89
mH/g kerosene = 10000/210 = 47.6
mH/g hydrogen = 285/2 = 142.5
mH ethanol = 1370/46 = 29.78
therefore its hydrogen
Part B
Density = 0.69g/mL
therefore mass octane = 80*1000 *0.69 = 55200
octane + oxygen --> carbon dioxide + water
im not gonna sit here n calculate the moles n stuff, but u basically find molar heat per gram and mole
Part c
use ur standard gas volume to figure out the volume of gas
so mH hydrogen = mH petrol
Please correct me if I'm wrong (which I might be)
mH (molar heat)/g for petrol = 5460/114 = 47.89
mH/g kerosene = 10000/210 = 47.6
mH/g hydrogen = 285/2 = 142.5
mH ethanol = 1370/46 = 29.78
therefore its hydrogen
Part B
Density = 0.69g/mL
therefore mass octane = 80*1000 *0.69 = 55200
octane + oxygen --> carbon dioxide + water
im not gonna sit here n calculate the moles n stuff, but u basically find molar heat per gram and mole
Part c
use ur standard gas volume to figure out the volume of gas
so mH hydrogen = mH petrol
Please correct me if I'm wrong (which I might be)
carlytse621
Member
a.) Hydrogen
b.)
G of petrol: 0.69 X 80000 = 55200g
n = m/MM = 55200/114 = 484.21 mol of petrol has combusted
energy released: 5460 X 484.21 = 2643786.6 kJ = 2.64 X 10^6 kJ
c.)
n required for 2.64 X 10^6 kJ : 2.64 X 10^6 / 285 = 9263.16 mol
V = n X MV = 9263.16 X 24.79
= 229633.74
= 2.30 X 10^5 L of hydrogen is needed
well....tell me if it is wrong....can't 100% sure it is right as i don't have the answer
b.)
G of petrol: 0.69 X 80000 = 55200g
n = m/MM = 55200/114 = 484.21 mol of petrol has combusted
energy released: 5460 X 484.21 = 2643786.6 kJ = 2.64 X 10^6 kJ
c.)
n required for 2.64 X 10^6 kJ : 2.64 X 10^6 / 285 = 9263.16 mol
V = n X MV = 9263.16 X 24.79
= 229633.74
= 2.30 X 10^5 L of hydrogen is needed
well....tell me if it is wrong....can't 100% sure it is right as i don't have the answer
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