# 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

Unknown..........

#### Explanation:

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

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

$\frac{1}{2} {N}_{2} \left(g\right) + \frac{3}{2} {H}_{2} \left(g\right) r i g h t \le f t h a r p \infty n s N {H}_{3} \left(g\right)$.....

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 $14 \cdot g$ of ammonia were isolated. We were NOT given the mass of dihydrogen or dinitrogen reactant, so we cannot address yield.

Jun 13, 2017

Assuming an attempt/intent to synthesize 1 mole $N {H}_{3}$, 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 ...

$\frac{3}{2} {H}_{2} \left(g\right)$ + $\frac{1}{2} {N}_{2} \left(g\right)$ => $N {H}_{3} \left(g\right)$
=> Theoretical yield from reaction = 1 mole $N {H}_{3} \left(g\right)$ = 17 grams $N {H}_{3} \left(g\right)$

Given Lab Yield = 14 grams $N {H}_{3} \left(g\right)$

=> $\text{%"Yield}$ = 14/17xx 100% = 82.4% w/w