Question

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

Answer 1

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.

Answer 2

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