Question #88b71

1 Answer
Apr 30, 2016

Answer:

#"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.