Question #64e35

The thing to remember about balanced chemical equations is that the coefficients added in front of each chemical substance tell you how many moles of said substances take part in the reaction.

In your case, the balanced chemical equation that describes the combustion of ammonia looks like this

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

This tells you that for every #color(red)(4)# moles of ammonia that take part in the reaction, the reaction consumes #5# moles of oxygen and produces #4# moles of nitric oxide and #color(blue)(6)# moles of water.

Now, you know that your reaction consumes #1.50# moles of ammonia. Since no mention of the number of moles of oxygen was made, you can assume that oxygen is in excess, i.e. that you have more moles of oxygen that you need.

You can use the #color(red)(4):color(blue)(6)# mole ratio that exists between ammonia and water to calculate the number of moles of water produced by the reaction

#1.50 color(red)(cancel(color(black)("moles NH"_3))) * (color(blue)(6)color(white)(a)"moles H"_2"O")/(color(red)(4)color(red)(cancel(color(black)("moles NH"_3)))) = color(darkgreen)(ul(color(black)("2.25 moles H"_2"O")))#

The answer is rounded to three sig figs.