Question #e811d

Start by writing a balanced chemical equation that describes this synthesis reaction

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

The thing to remember about balanced chemical equations is that the coefficients added in front of a chemical species represent the number of moles of said chemical species needed in order for the reaction to take place.

In this case, you need #1# mole of hydrogen gas for every #color(blue)(3)# moles of hydrogen gas in order to produce #color(red)(2)# moles of ammonia.

Now, these mole ratios are true regardless of how many moles of a particular chemical species you have. In other words, you will have

#"number of moles of N"_2/"number of moles of NH"_3 = 1/color(red)(2)#

Similarly, you will also have

#"number of moles of H"_2/"number of moles of N"_2 = color(blue)(3)/1#

and

#"number of moles of H"_2/"number of moles of NH"_3 = color(blue)(3)/color(red)(2)#

In your case, you know that #0.85# moles of ammonia were produced by the reaction. Since the #1:color(red)(2)# mole ratio that exists between nitrogen gas and ammonia tells you that

#"no. of moles of N"_2 = 1/color(red)(2) * "no. of moles of NH"_3#

you can say that the reaction consumed

#"no. of moles of N"_2 = 1/color(red)(2) * "0.85 moles" = color(darkgreen)(ul(color(black)("0.43 moles N"_2)))#

The answer must be rounded to two sig figs, the number of sig figs you have for the number of moles of ammonia produced by the reaction.