# How much energy is released when 14.0 g of carbon monoxide are completely combusted?

## $2 C {O}_{\left(g\right)} + {O}_{2 \left(g\right)} \to 2 C {O}_{2 \left(g\right)} \text{ " DeltaH = -"560 kJ}$

Mar 10, 2016

The reaction will give off $\text{140 kJ}$ of heat.

#### Explanation:

The problem provides you with the thermochemical equation for the combustion of carbon monoxide, $\text{CO}$

$\textcolor{red}{2} \text{CO"_text(2(g]) + "O"_text(2(g]) -> 2"CO"_text(2(g])" " DeltaH_text(rxn) = -"560 kJ}$

In essence, the thermochemical equation tells you the enthalpy change of reaction, $\Delta {H}_{\text{rxn}}$, for when the reaction consumes the number of moles of reactants and produces the number of moles of product(s) that act as stoichiometric coefficients in the balanced chemical equation.

In this case, you can say that when $\textcolor{red}{2}$ moles of carbon monoxide react with $1$ mole of oxygen gas, $2$ moles of carbon dioxide are produced and the enthalpy change of reaction is $\Delta {H}_{\text{rxn" = -"560 kJ}}$.

In other words, when $\textcolor{red}{2}$ moles of carbon monoxide undergo combustion, $\text{560 kJ}$" of heat are being given off, hence the minus sign used for $\Delta {H}_{\text{rxn}}$.

So, you know that the reaction gives off $\text{560 kJ}$ of heat when two moles of carbon monoxide react. Use the compound's molar mass to determine how many moles you get in that $\text{14.0-g}$ sample

14.0color(red)(cancel(color(black)("g"))) * overbrace("1 mole CO"/(28.0101color(red)(cancel(color(black)("g")))))^(color(brown)("molar mass of CO")) = "0.4998 moles CO"

You can now use the known enthalpy change of reaction for when two moles of carbon monoxide undergo combustion as a conversion factor to figure out how much heat is being given off when $0.4998$ moles react

0.4998color(red)(cancel(color(black)("moles CO"))) * overbrace("560 kJ"/(color(red)(2)color(red)(cancel(color(black)("moles CO")))))^(color(purple)("the known"color(white)(a) DeltaH_"rxn")) = "139.94 kJ"

Rounded to two sig figs, the number of sig figs you have for $\Delta {H}_{\text{rxn}}$, the answer will be

"heat given off" = color(green)(|bar(ul(color(white)(a/a)"140 kJ"color(white)(a/a)|)))

This is equivalent to saying that the enthalpy change of reaction when $\text{14.0 g}$ of carbon monoxide undergo cumbustion is equal to

DeltaH_"rxn for 14.0 g" = color(green)(|bar(ul(color(white)(a/a)-"140 kJ"color(white)(a/a)|)))

Remember, the minus sign denotes heat given off.

Here is a video which shows another example of how to complete a question like this one. The video discusses the combusion of butane.