If #14*g# of ammonia was isolated from a reaction between #N_2#, and #H_2#, what is the percentage yield?

2 Answers
Jun 13, 2017

Answer:

Unknown..........

Explanation:

#"% yield"="moles of products"/"moles of reactants"xx100%#

And thus quotient would have to be appropriately modified for the given reaction:

#1/2N_2(g) + 3/2H_2(g) rightleftharpoons NH_3(g)#.....

i.e. #"yield"="Moles of ammonia"/(1/2xx"moles of dinitrogen")# OR

#"yield"="Moles of ammonia"/(3/2xx"moles of dihydrogen")#

Now we were given that #14*g# of ammonia were isolated. We were NOT given the mass of dihydrogen or dinitrogen reactant, so we cannot address yield.

Jun 13, 2017

Answer:

Assuming an attempt/intent to synthesize 1 mole #NH_3#, then %yield = #82.4% Yield# w/w

Explanation:

#%Yield = ("Lab Yield"/"Theoritical Yield")xx100%#

For the sake of showing how %Yield is calculated, assume the intent is to synthesize 1 mole ammonia. Then the theoretical Yield from equation stoichiometry ...

#3/2H_2(g)# + #1/2N_2(g)# => #NH_3(g)#
=> Theoretical yield from reaction = 1 mole #NH_3(g)# = 17 grams #NH_3(g)#

Given Lab Yield = 14 grams #NH_3(g)#

=> #"%"Yield"# = #14/17xx 100% = 82.4%# w/w