# 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?

## $C {H}_{4} \left(g\right) + 2 {O}_{2} \left(g\right) \to C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(g\right)$ $\Delta H = - \text{802.3 kJ/mol}$

Jul 18, 2016

$\text{11900 kJ}$

#### Explanation:

The problem provides you with the thermochemical equation that describes the combustion of methane, ${\text{CH}}_{4}$

${\text{CH"_ (4(g)) + 2"O"_ (2(g)) -> "CO"_ (2(g)) + 2"H"_ 2"O"_ ((g))" "DeltaH = - "802.3 kJ mol}}^{- 1}$

The enthalpy change of combustion, given here as $\Delta H$, 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

$\Delta H = - {\text{802.3 kJ mol}}^{- 1}$

suggests that the combustion of one mole of methane gives off, hence the minus sign, $\text{802.3 kJ}$ 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 $\text{802.3 kJ}$ 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, $\Delta {H}_{\text{rxn}}$, when $\text{237 g}$ of methane undergo combustion is

$\Delta {H}_{\text{rxn" = -"11900 kJ}}$

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