Question #3fa48

It is absolutely vital that you work with a balanced chemical equation, so 8always* make sure that your equation is balanced before doing any stoichiometric calculations.

In this case, the balanced chemical equation is

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

Notice that you have a #color(blue)(3):color(purple)(2)# mole ratio between hydrogen gas and ammonia. This ratio tells you that the number of moles of hydrogen gas that react will always be in a #3/2# ratio with the number of moles of ammonia produced.

Now, the problem tells you that you start with #21.0# moles of hydrogen gas. Because nitrogen gas is in excess, all the moles of hydrogen gas will react.

This will get you

#21.0 color(red)(cancel(color(black)("moles H"_2))) * (color(purple)(2)color(white)(a)"moles NH"_3)/(color(blue)(3)color(red)(cancel(color(black)("moles H"_2)))) = "14.0 moles NH"_3#

Notice that the problem gives you moles of hydrogen gas and asks for moles of ammonia, so there's absolutely no need to include any molar masses in your calculations.

Moreover, molar masses express the mass of exactly one mole of a substance. In this case, the molar mass of hydrogen gas would be #"2.016 g/mol"#, meaning that you have

#"2.016 g H"_2/"1 mole H"_2#

That is your conversion factor. The same goes for nitrogen gas, which has a molar mass of #"28.01 g/mol"#. This conversion factor will be

#"28.01 g N"_2/"1 mole N"_2#

In this regard, I don't know what these conversion factors

#"1 mole H"_2/"2.016 mol H"_2" "# #" " "28.02 mol N"_2/"1 mole N"_2" "# #" " "588.42 N"_2/6.048#

are supposed to represent. Remember, the molar mass represents the mass of one mole of a substance, which is why you must always have a mass in there when you use a molar mass as a conversion factor.