Question #a19b3

The balanced chemical equation that describes this reaction tells you that you need #1# mole of nitrogen gas to produce #2# moles of ammonia.

#"N"_ (2(g)) + 3"H"_ (2(g)) -> 2"NH"_ (3(g))#

So right from the start, you can use the molar mass of nitrogen gas and the molar mass of ammonia to say that every

#1 color(red)(cancel(color(black)("mole N"_2))) * "28.0134 g"/(1color(red)(cancel(color(black)("mole N"_2)))) = "28.0134 g"#

of nitrogen gas that react, the reaction produces

#2 color(red)(cancel(color(black)("moles NH"_3))) * "17.031 g"/(1color(red)(cancel(color(black)("mole NH"_3)))) = "34.062 g"#

of ammonia. So if you know that the reaction produced #"7.5 g"# of ammonia, you can use this gram ratio to determine the mass of nitrogen gas that took palce in the reaction, assuming, of course, that the reaction has a #100%# yield.

#7.5 color(red)(cancel(color(black)("g NH"_3))) * "28.0134 g N"_2/(34.062color(red)(cancel(color(black)("g NH"_3)))) = color(darkgreen)(ul(color(black)("6.2 g N"_2)))#

The answer is rounded to two sig figs.