Question #bb0ad

The balanced chemical equation that describes your reaction looks like this

#4"NH"_ (3(g)) + 5"O"_ (2(g)) -> 4"NO"_ ((g)) + 6"H"_ 2"O"_ ((l))#

Notice that the reaction consumes #5# moles of oxygen gas for every #4# moles of ammonia that takes part in the reaction. In other words, the number of moles of ammonia and the number of moles of oxygen gas that take part in the reaction will always be in a #4:5# ratio.

In your case, you know that #1# mole of ammonia reacts with #1# mole of oxygen gas. The fact that you have equal numbers of moles of the two reactants and a #4:5# mole ratio between them tells you that you're dealing with a limiting reagent.

More specifically, you know that #1# mole of ammonia would require

#1 color(red)(cancel(color(black)("mole NH"_3))) * overbrace("5 moles O"_2/(4color(red)(cancel(color(black)("moles NH"_3)))))^(color(blue)("given by the balanced chemical equation")) = "1.25 moles O"_2#

Since you only have #1# mole of oxygen gas available, you can conclude that oxygen gas will act as the limiting reagent, i.e. it will be completely consumed by the reaction before all the moles of ammonia will get the chance to react.

The reaction will consume #1# mole of oxygen gas and only

#1 color(red)(cancel(color(black)("mole O"_2))) * "4 moles NH"_3/(5color(red)(cancel(color(black)("moles O"_2)))) = "0.8 moles NH"_3#

The remaining

#overbrace("1 mole NH"_3 )^(color(blue)("what you start with"))- overbrace("0.8 moles NH"_3)^(color(blue)("consumed by the raction")) = overbrace("0.2 moles NH"_3)^(color(blue)("in excess"))#

will not take part in the reaction.

Moreover, the reaction will produce

#1 color(red)(cancel(color(black)("mole O"_2))) * "4 moles NO"/(5color(red)(cancel(color(black)("moles O"_2)))) = "0.8 moles NO"#

and

#1 color(red)(cancel(color(black)("mole O"_2))) * ("6 moles H"_2"O")/(5color(red)(cancel(color(black)("moles O"_2)))) = "1.2 moles H"_2"O"#