How many kilojoules of heat are released when 0.72 mole of oxygen gas are used to combust methane?

Answer:

459.6kj

Explanation:

the chemical equation =

3mn(s) + 2CO2(g) --- Mn3O4(s)

we have change in H to be -1378.83

we are required to find the amount of heat that is liberated for a mole of manganese.

3 moles of manganese = -1378.83

1 mole of manganese = ?

when we cross multiply, we will have a ratio that we will use to find this amount

1378.83/3 = 459.6kj

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.

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"_ ((l))" "DeltaH = -"890.3 kJ mol"^(-1)#

Notice that the thermochemical equation includes the enthalpy change of combustion, which as you can see is given in kilojoules per mole, #"kJ mol"^(-1)#.

This tells you that when one mole of methane undergoes combustion, #"890.3 kJ"# of heat are given off. This is equivalent to saying that the enthalpy change for the combustion of one mole of methane is

#DeltaH = -"890.3 kJ"#

The minus sign is used to denote heat given off.

So, you know the enthalpy change of reaction associated with the combustion of one mole of methane. Use methane's molar mass to calculate how many moles you have in your sample

#7.00 color(red)(cancel(color(black)("g"))) 8 "1 mole CH"_4/(16.04color(red)(cancel(color(black)("g")))) = "0.4364 moles CH"_4#

Use the enthalpy change of combustion to find the heat given off when #0.4364# moles of methane undergo combustion

#0.4364 color(red)(cancel(color(black)("moles CH"_4))) * "890.3 kJ"/(1color(red)(cancel(color(black)("mole CH"_4)))) = "389 kJ"#

Now, the enthalpy change of reaction, #DeltaH#, will be

#DeltaH = color(green)(|bar(ul(color(white)(a/a)color(black)(-"389 kJ")color(white)(a/a)|)))#

Once again, keep in mind that the minus sign is needed because the heat is being given off.

The answer is rounded to three sig figs.

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