Question #0314e

Nov 25, 2016

$\text{13.35 g}$

Explanation:

You could start by converting the heat of combustion, also known as the enthalpy of combustion, $\Delta {H}_{\text{comb}}$, from kilojoules per mole to kilojoules per gram, since that would be more useful in determining the number of grams needed to produce $\text{296.20 kJ}$.

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

$\Delta {H}_{\text{comb" = - "1170.0 kJ mol}}^{- 1}$

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

when $\text{1 mole}$ of this compound undergoes combustion, exactly $\text{1170.0 kJ}$ of heat are being given off

Since you know that the compound has a molar mass of ${\text{52.73 g mol}}^{- 1}$, you can say that the enthalpy of combustion is equivalent to

$- 1170.0 {\text{kJ"/color(red)(cancel(color(black)("mol"))) * (1color(red)(cancel(color(black)("mol"))))/"52.73 g" = -"22.1885 kJ g}}^{- 1}$

This means that

when $\text{1 gram}$ of this compound undergoes combustion, exactly $\text{22.1885 kJ}$ 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

$296.20 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{kJ"))) * "1 g"/(22.1885color(red)(cancel(color(black)("kJ")))) = color(darkgreen)(ul(color(black)("13.35 g}}}}$

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