(Q&A)I need to calculate volume of C7H16 required to boil 36L of water with 15% efficiency

Q:-
std enthalpy is -224.4 kJ
density is .78g/mol

36,000g * 4.184 * 75 = 11296.8 kJ required to heat the water

which leaves (224.4*.15)x = 11296.8 or x = 335.62 moles of C7H16

335.62 moles * 100 g/mol = 33562 g / .78 = 43027.56 mL of C7H16 required to heat the water...

Except the answer 43027.56 is not right...

Here is the problem:
When backpacking in the wilderness, hikers often boil water to sterilize it for drinking.
Suppose that you are planning a backpacking trip and will need to boil 36 L of water for your group.

What volume of fuel should you bring? Assume each of the following:
the fuel has an average formula of C7H16 ; 15% of the heat generated from combustion goes
to heat the water (the rest is lost to the surroundings);
the density of the fuel is 0.78 g/mL; the initial temperature
of the water is 25.0C; and the standard enthalpy of
formation of C7H16 is -224.4 kJ/mol..

i'm still missing something subtle... I come out to 43000 mL of C7H16 each time....

A:-
Your problem is the you supplied the dHf of heptane and NOT the heat of combustion!

-224.4 kJ/mole vs -44.566MJ/kg is a BIG difference.
I thought 224 kJ for combustion was small but did not check it out earlier.

Let UNITs guide you; always USE THEM in your calculation to prevent errors

Firstly, the eqn: C7H16 + 11O2 -----> 7CO2 + 8H2O + heat

dH comb/ mole C7H16 = 44.566 MJ /kg C7H16 = 44.566 kJ/g

dH comb J /ml C7H16 = 44.566kJ/ g * 0.78 g/ml = ?? 34.76 kJ/ml

heat needed, if 100% eff = 36000 g * 4.184 J/gC * 75 C = ??
11.3 MJ

if 15% eff then heat needed = 100% eff/15% * heat needed if 100% = 75.3 MJ

vol of C7H16 needed = heat needed / (+dH comb / ml C7H16) = 75.3 MJ / 34.76 kJ/ml ?? 2167 ml

liters C7H16 needed = vol ml / 1000ml/liter = ?? 2.17 liters

1 liter = 0.26 US gal; so 2.17 liters * 0.26 gal/liter = 0.56 gal

Plug and SOLVE

Basic mathematics is a prerequisite to chemistry – I just try to help you with the methodology of solving the problem.