# Concerning the reaction C + O_2 -> CO_2, if the ΔH° = -393 kJ/mol, how many grams of C must be burned to release 275 kJ of heat?

Dec 8, 2015

$\text{8.40 g}$

#### Explanation:

Notice that the standard enthalpy change of reaction, $\Delta {H}^{\circ}$, is given to you in kilojoules per mole, which means that it corresponds to the formation of one mole of carbon dioxide.

${\text{C"_text((s]) + "O"_text(2(g]) -> "CO}}_{\textrm{2 \left(g\right]}}$

More specifically, when one mole of carbon reacts with one mole of oxygen gas, one mole of carbon dioxide is produced and $\text{393 kJ}$ of heat are given off.

Remember, a negative enthalpy change of reaction tells you that heat is being given off, i.e. the reaction is exothermic.

Now, in order to determine how much carbon is needed in order for the reaction to give off $\text{275 kJ}$ of heat, use the fact that one mole of carbon is needed to give off $\text{393 kJ}$ of heat.

275 color(red)(cancel(color(black)("kJ"))) * "1 mole C"/(393 color(red)(cancel(color(black)("kJ")))) = "0.6997 moles C"

To see how many grams of carbon would contain this many moles, use carbon's molar mass

0.6997 color(red)(cancel(color(black)("moles C"))) * "12.011 g"/(1color(red)(cancel(color(black)("mole C")))) = color(green)("8.40 g")

So, when $8.40$ grams of carbon react with enough oxygen gas, the reaction will give off $\text{275 kJ}$ of heat. This is equivalent to having a standard enthalpy change of reaction equal to

$\Delta {H}^{\circ} = - \text{275 kJ}$