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?

1 Answer
Truong-Son N.
Apr 4, 2017

"2.76 g NO"

If the units were more clear, we would realize that the "114.14 kJ" is really DeltaH_"rxn"^@ = -"114.14 kJ/"ulbb"mol".
That is, we have an exothermic reaction that releases "114.14 kJ" for every mol of "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 "mol"s of "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 = (5.25/57.07) "mols"

= "0.0920 mols NO"

Therefore, the mass of "NO" is simply gotten from its molar mass.

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

= color(blue)("2.76 g NO")

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