74. The equation for the combustion of #CH_4# (the main component of natural gas) is shown below. How much heat is produced by the complete combustion of 237 g of #CH_4#?

#CH_4 (g) + 2O_2 (g) -> CO_2 (g) + 2H_2O (g)#
#DeltaH = -"802.3 kJ/mol"#

1 Answer
Jul 18, 2016

Answer:

#"11900 kJ"#

Explanation:

The problem provides you with the thermochemical equation that describes the combustion of methane, #"CH"_4#

#"CH"_ (4(g)) + 2"O"_ (2(g)) -> "CO"_ (2(g)) + 2"H"_ 2"O"_ ((g))" "DeltaH = - "802.3 kJ mol"^(-1)#

The enthalpy change of combustion, given here as #DeltaH#, tells you how much heat is either absorbed or released by the combustion of one mole of a substance.

In your case, the enthalpy change of combustion

#DeltaH = -"802.3 kJ mol"^(-1)#

suggests that the combustion of one mole of methane gives off, hence the minus sign, #"802.3 kJ"# of heat.

Your strategy here will be to use the molar mass of methane to convert your sample from grams to moles

#237 color(red)(cancel(color(black)("g"))) * "1 mole CH"_4/(16.04color(red)(cancel(color(black)("g")))) = "14.776 moles CH"_4#

Since you know that #1# mole produces #"802.3 kJ"# of heat upon combustion, you can say that #14.776# moles will produce

#14.776 color(red)(cancel(color(black)("moles CH"_4))) * overbrace("802.3 kJ"/(1color(red)(cancel(color(black)("mole CH"_4)))))^(color(blue)(= DeltaH)) = "11854.8 kJ"#

Rounded to three sig figs, the answer will be

#"heat produced" = color(green)(|bar(ul(color(white)(a/a)color(black)("11900 kJ")color(white)(a/a)|)))#

This is equivalent to saying that the enthalpy change of reaction, #DeltaH_"rxn"#, when #"237 g"# of methane undergo combustion is

#DeltaH_"rxn" = -"11900 kJ"#

Keep in mind that the minus sign is used to symbolize heat given off.