Question

Question #88b71

Answer 1

`"8.44 g CH"_4`

Explanation:

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

`"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, `"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 `"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 `"16.04 g mol"^(-1)`, which means that one mole of methane has a mass of `"16.04 g"`. This means that the reaction given to you gives off `"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 `"16.04 g"` of methane react, it follows that `"422 kJ"` of heat will be given off by

`422 color(red)(cancel(color(black)("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"_4color(white)(a/a)|)))`

The answer is rounded to two sig figs.

So, when `"8.44 g"` of methane undergo combustion, `"422 kJ"` of heat are being given off. This is equivalent to saying that when `"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.