# If the heat of combustion for a specific compound is -1160.0 kJ/mol and its molar mass is 84.45 g/mol, how many grams of this compound must you burn to release 427.20 kJ of heat?

Nov 29, 2015

$\text{31.10 g}$

#### Explanation:

Your strategy here is to use the compound's molar heat of combustion and its molar mass to determine the enthalpy change of combustion per gram.

So, you know that burning one mole of this compound will releases $\text{1160.0 kJ}$ of heat. This will be equivalent to

$1160.0 \text{kJ"/color(red)(cancel(color(black)("mol"))) * (1color(red)(cancel(color(black)("mole"))))/"84.45 g" = "13.736 kJ/g}$

Now, if your compound gives of $\text{13.736 kJ}$ per gram upon combustion, it follows that $\text{427.20 kJ}$ of heat will be given off by the combustion of

427.20color(red)(cancel(color(black)("kJ"))) * "1 gram"/(13.736color(red)(cancel(color(black)("kJ")))) = color(green)("31.10 g")

Therefore, burning $\text{31.10 g}$ of this compound will result in $\text{427.20 kJ}$ of heat being released. This is of course equivalent to saying that the enthalpy change of combustion when $\text{31.10 g}$ of the compound undergo combustion is

$\Delta {H}_{\text{comb" = -"427.20 kJ}}$

Here the minus sign is used to designate heat given off.