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")#

Related questions
See all questions in Enthalpy
Impact of this question
1705 views around the world
You can reuse this answer
Creative Commons License