In reaction 2 NH3 (g) → 3 H2 (g) + N2 (g), if 12.0 g of ammonia produced 1.87 g of hydrogen, what was the percent yield?

Quote from: Harini on September 18, 2022, 01:40:36 PM
In reaction 2 NH3 (g) → 3 H2 (g) + N2 (g), if 12.0 g of ammonia produced 1.87 g of hydrogen, what was the percent yield?

In this question - there is only one reactant and its amount is given.
This question doesn't have two reactants so it is not a question of limiting reactant. A question in which we need to determine limiting reactant  has more than one reactant and their amounts are given.
In the given question you need to calculate theoretical yield first ?.start with amount of ammonia given .Here is the planning -
g of NH₃ ----> moles of ammonia ----> moles of H₂ -----> g H₂ (theoretical yield)
Grams to moles you can use ratio that 1mol = molar mass
moles to moles you can use molar ratio from balanced equation
like in this case 2 moles of NH₃  = 3 moles of H₂
So what is theoretical yield -it is the amount calculated using given amount.
Actual yield is the experimental yield which is given in this question as 1.87 g H2
percent yield = (Actual yield /Theoretical yield  )100%
Try this and use dimensional method.
Share your work here.

My work
12.0g x 1mole of ammonia / 17.031g/mol of NH3 x3H2/2NH3x2g H2 = 2.1g of H2 = theoretical yield
(1.87/2.10) x 100% = 89% yield

Correct