How much ammonia should be produced given that dinitrogen reacts quantitatively with a #2.57xx10^-4*g# mass of dihydrogen gas?

1 Answer
Sep 30, 2017

Let's see........

Explanation:

First we need a stoichiometric equation.....

#1/2N_2(g) + 3/2H_2(g) rightleftharpoonsNH_3(g)#

We gots a mass of #2.57xx10^-4*g# with respect to dihydrogen gas....and thus a molar quantity of ,,,,

#(2.57xx10^-4*g)/(2.016*g*mol^-1)=1.275xx10^-4*mol#.

And given the stoichiometry, there should be #2/3xx1.275xx10^-4*mol=8.50xx10^-5*mol# with respect to ammonia. (Of course, we assume (unreasonably!) quantitative yield....)

And we multiply this molar quantity by the #"Avocado number"# to get the number of molecules.....

#8.50xx10^-5*molxx6.022xx10^-23*mol^-1-=5.12xx10^19# #"ammonia molecules."#

It is all too easy to make an arithmetic mistake with these answers...