Question #0f579

In order to be able to answer this question, you need to know how the balanced chemical equation for this synthesis reaction looks like.

Nitrogen gas and hydrogen gas will react to produce ammonia, #"NH"_3#, according to the balanced chemical equation

#"N"_text(2(g]) + color(blue)(3)"H"_text(2(g]) -> color(red)(2)"NH"_text(3(g])#

Now, the mole ratios that exist between the species that take part in the reaction will help you determine how many moles of ammonia will be produced when #0.5# moles of nitrogen gas and of hydrogen gas react.

As you can see, the #1:color(blue)(3)# mole ratio that exists between nitrogen and hydrogen tells you that the reaction consumes three times more moles of hydrogen than moles of nitrogen.

This means that if you start with equal number of moles of both reactants, the hydrogen gas will act as a limiting reagent, i.e. it will limit the amount of nitrogen that takes part in the reaction.

More specifically, only

#0.5color(red)(cancel(color(black)("moles H"_2))) * "1 mole N"_2/(color(blue)(3)color(red)(cancel(color(black)("moles H"_2)))) = 1/6 " moles N"_2#

will actually react, the remaining

#0.5 - 1/6 = 2/3" moles N"_2#

will be in excess.

Now look at the mole ratio that exists between nitrogen gas and ammonia. If #1/6# moles of nitrogen gas take part in the reaction, then you can say that you have

#1/6color(red)(cancel(color(black)("moles N"_2))) * (color(red)(2)" moles NH"_3)/(1color(red)(cancel(color(black)("mole N"_2)))) = 1/3 = "0.33 moles NH"_3#

Rounded to one sig fig, the answer will be

#n_(NH_3) = color(green)("0.3 moles NH"_3)#