# Combustion of paraffin

Find the heat of combustion for 1 mol of $\ce {C21H44}$ in the reaction where some wax is oxidized to elemental C (soot) and water vapor. The Heat of Formation of $\ce {C21H44}$ is -476 kJ/mol.

When it says some, does that necessarily change the molar ratio? I thought the balanced equation would be:

$\ce {C21H44 + 11O2 -> 21C (graphite) + 22H2O}$

$\ce {O2}$ and elemental carbon are zero for heat of formation

22(-241.818 kJ/mol) - (-476 kJ/mol) = -4843.996 kJ/mol

Is this right?

Thanks

2. is asking about a reaction in which part of the paraffin is being oxidized to C and part is being oxidized to CO2, the usual reaction that occurs. Though that is more likely to occur in real life, in my opinion, it also is impossible to solve without further information. The best you could do would be to set a lower limit (the value you've already found) and an upper limit, the heat of reaction for $\ce{C21H44 + 32 O2 -> 21 CO2 + 22H2O}$.