Question

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"`

Answer 1

`"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.