year 11 chemsitry question. (1 Viewer)

[girlworld_club]

calculate the heat released per gram from the combustion of each of the following fuels (heats of combustion in KJ/mol are given in brackets)

Hydrogen (285)
Propane (2220)

please show full working.

 

[someth1ng]

deltaH/m=deltaH/n*(n/m)
OR
deltaH/m=deltaH/n*(1/M)

deltaH/m(H2)=deltaH(H2)/n(H2)*(1/M(H2))
deltaH(H2)/m(H2)=285*1/2
deltaH(H2)/m(H2)=142.5kJ/gram

deltaH(C3H8)/m(C3H8)=deltaH(C3H8)/n(C3H8)*(1/M(C3H8))
deltaH(C3H8)/m(C3H8)=2220*1/(3*12+8*1)
deltaH(C3H8)/m(C3H8)=50.45kJ/gram

 

[girlworld_club]

there is a simpler way of solving this question..i haven't encountered this formulae yet.

 

[wannabesurgeon]

Hydrogen(285)

Divide 285 by the molar mass of Hydrogen which is 1.008

(kj/mol)/(grams/mols)= Kj(mols)/(grams x mols)

It gives you 282.738 kJ/g

For Propane(2220) C3H8 - MM is 44.094

Divide 2220 by 44.094

Gives you 50.347 kJ/g which if you round it off to 2 decimal places it is the same as someth1ngs's answer..

 

[Girls]

Hydrogen(285)

Divide 285 by the molar mass of Hydrogen which is 1.008

(kj/mol)/(grams/mols)= Kj(mols)/(grams x mols)

It gives you 282.738 kJ/g

I'm going to guess that the question meant H2 as a fuel, rather than just an atom of hydrogen.