# Question 88b71

Apr 30, 2016

${\text{8.44 g CH}}_{4}$

#### Explanation:

Start by writing down the thermochemical equation that describes the combustion of methane

$\text{CH"_ (4(g)) + 2"O"_ (2(g)) -> "CO"_ (2(g)) + 2"H"_ 2"O"_ ((g))" "DeltaH_"rxn"^@ = -"802.3 kJ}$

This equation tells you that when one mole of methane undergoes combustion, $\text{802.3 kJ}$ of heat are being given off by the reaction.

Since the problem asks for the mass of methane needed to give off $\text{422 kJ}$ of heat, you can use methane's molar mass to express the heat given off when one mole undergoes combustion.

Methane has a molar mass of ${\text{16.04 g mol}}^{- 1}$, which means that one mole of methane has a mass of $\text{16.04 g}$. This means that the reaction given to you gives off $\text{802.3 kJ}$ of heat when

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

of methane undergo combustion. So, if this much heat is given off when $\text{16.04 g}$ of methane react, it follows that $\text{422 kJ}$ of heat will be given off by

$422 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{kJ heat"))) * "16.04 g CH"_4/(802.3 color(red)(cancel(color(black)("kJ heat")))) = color(green)(|bar(ul(color(white)(a/a)"8.44 g CH}}_{4} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to two sig figs.

So, when $\text{8.44 g}$ of methane undergo combustion, $\text{422 kJ}$ of heat are being given off. This is equivalent to saying that when $\text{8.44 g}$ of methane react, the standard enthalpy change of reaction is equal to

DeltaH_("rxn 8.44 g CH"_4)^@ = -"422 kJ"#

The minus sign symbolizes heat lost.