What mass of methane combusted?

I got about #"1.19 g CH"_4#. However, I would actually have thought that #DeltaH_c^@ = -"802.5 kJ/mol"#, in a truly constant pressure process.

Just a sidenote, but that value uses "#DeltaH_(f,H_2O(l))^@#" #= -"285.8 kJ/mol"#, so it was done in a bomb calorimeter (at constant volume, not constant pressure). In such a scenario, #q = DeltaU_c^@# (not #DeltaH_c^@#), where #U# is the internal energy.

Anyways, since we know the volume of water affected, we know the heat flow is:

#q = mC_PDeltaT#

where #m# is the mass of solution, #C_P# is the specific heat capacity at constant atmospheric pressure (not volume), and #DeltaT# is the change in temperature.

Here we assume that the density of water is practically #"1 g/mL"# or #"1 kg/L"#, so:

#q ~~ 2.40 xx 10^3 "g" cdot "4.184 J/g"^@ "C" cdot (24.8^@ "C" - 18.2^@ "C")#

#=# #+"66274.56 J"# #=# #"66.275 kJ"#

as this heat was absorbed in this endothermic process.

This heat involved could have come from some source, and we are presuming the source is the combustion of a certain mass of methane.

#"66.275 kJ"/("??? mol CH"_4) = "890.5 kJ"/"1 mol CH"_4#

They must be equal, as long as energy is conserved, since the heat should be released from the combustion and absorbed into the water.

(This number should also be less than #1#, because energy in units without "#"/mol"#" is extensive, meaning less mols of combusted #"CH"_4# gives less heat out.)

#"mol CH"_4 = 66.275 cancel"kJ" xx "1 mol CH"_4/(890.5 cancel"kJ")#

#= "0.0744 mols CH"_4#

Thus, heating #"2.40 L"# of water up from #18.2^@ "C"# to #24.8^@ "C"# required combustion of about

#color(blue)(m_(CH_4(g))) = 0.0744 cancel("mols CH"_4) xx ("16.0426 g CH"_4)/cancel("1 mol CH"_4)#

#= color(blue)("1.19 g CH"_4)#