What volume of nitrogen monoxide is produced when #"146 L"# of ammonia react according to the following reaction? (All gases are at the same temperature and pressure.)

Start by converting the word equation to a balanced chemical equation.

You know that ammonia, #"NH"_3#, reacts with oxygen gas, #"O"_2#, to produce nitric oxide, #"NO"#, which is just another name for nitrogen monoxide, and water vapor, so you can write a balanced chemical equation that looks like this:

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

Now, notice that in order for the reaction to produce #4# moles of nitric oxide, it must consume #4# moles of ammonia.

This means that if you have enough moles of oxygen gas available, the number of moles of nitric oxide produced by the reaction will be equal to the number of moles of ammonia consumed.

Since all gases are kept under the same conditions for pressure and temperature, you can express this mole ratio in terms of the volumes occupied by each gas.

In other words, when the reaction consumes #4# liters of ammonia ir produces #4# liters of nitric oxide, provided, of course, that you have enough oxygen gas available, i.e. #5# liters.

So you can say that when oxygen gas is in excess, as you have here, when the reaction consumes #"146 L"# of ammonia and, it produces

#146 color(red)(cancel(color(black)("L NH"_3))) * "4 L NO"/(4color(red)(cancel(color(black)("L NH"_3)))) = color(darkgreen)(ul(color(black)("146 L NO")))#

The answer is rounded to three sig figs.