# For the reaction "N"_2(g) + "O"_2(g) -> 2"NO"(g) + "114.14 kJ", what mass of "NO" would need to be produced in order to release "5.25 kJ" of heat?

Apr 4, 2017

$\text{2.76 g NO}$

If the units were more clear, we would realize that the $\text{114.14 kJ}$ is really $\Delta {H}_{\text{rxn"^@ = -"114.14 kJ/"ulbb"mol}}$.
That is, we have an exothermic reaction that releases $\text{114.14 kJ}$ for every mol of ${\text{O}}_{2}$ that is used (since it is the substance with a stoichiometric coefficient of $1$).

So, you can set up the following expression:

"114.14 kJ"/("mol O"_2) = "5.25 kJ"/("x mol NO")

What we can then do is rewrite the lefthand side to use $\text{mol}$s of $\text{NO}$ instead.

"114.14 kJ"/cancel("mol O"_2) xx cancel("1 mol O"_2)/("2 mol NO")

= "57.07 kJ"/("mol NO")

Thus, we now have:

"57.07 kJ"/("mol NO") = "5.25 kJ"/("x mol NO")

Solve for $x$:

$x = \left(\frac{5.25}{57.07}\right) \text{mols}$

$=$ $\text{0.0920 mols NO}$

Therefore, the mass of $\text{NO}$ is simply gotten from its molar mass.

0.0920 cancel"mols NO" xx (14.007+15.999 "g NO")/(cancel"mols NO")

$=$ $\textcolor{b l u e}{\text{2.76 g NO}}$