Question

Question #0314e

Answer 1

`"13.35 g"`

Explanation:

You could start by converting the heat of combustion, also known as the enthalpy of combustion, `DeltaH_"comb"`, from kilojoules per mole to kilojoules per gram, since that would be more useful in determining the number of grams needed to produce `"296.20 kJ"`.

So, you know that the enthalpy of combustion for a given compound is equal to

`DeltaH_"comb" = - "1170.0 kJ mol"^(-1)`

The minus sign is used here to denote heat given off by the reaction, so you could say that

when `"1 mole"` of this compound undergoes combustion, exactly `"1170.0 kJ"` of heat are being given off

Since you know that the compound has a molar mass of `"52.73 g mol"^(-1)`, you can say that the enthalpy of combustion is equivalent to

`-1170.0 "kJ"/color(red)(cancel(color(black)("mol"))) * (1color(red)(cancel(color(black)("mol"))))/"52.73 g" = -"22.1885 kJ g"^(-1)`

This means that

when `"1 gram"` of this compound undergoes combustion, exactly `"22.1885 kJ"` of heat are being given off

Now all you have to do is use the enthalpy of combustion as a conversion factor to go from heat given off to grams

`296.20color(red)(cancel(color(black)("kJ"))) * "1 g"/(22.1885color(red)(cancel(color(black)("kJ")))) = color(darkgreen)(ul(color(black)("13.35 g")))`

The answer is rounded to four sig figs, the number of sig figs you have for the molar mass of the compound.