Question

3h2(g)+n2(g)→2nh3(g) what is the theoretical yield of ammonia, in kilograms, that we can synthesize from 5.25 kg of h2 and 32.7 kg of n2?

Answer 1

Answer:
29.75 Kg of NH₃

Solution:

In order to calculate the theoretical yield, first we will identify the limiting reactant.
According to equation,

6 g (3 moles) H₂ requires = 28 g (1 mole) N₂
So,
5250 g H₂ will require = X g of N₂

Solving for X,
X = (5250 g × 28 g) ÷ 6 g

X = 25433 g of N₂

Hence, it is found that H₂ is the limiting reactant because N₂ is provided in excess (32700 g). Therefore,
As,
6 g (3 mole) H₂ produced = 34 g ( 2 moles) of NH₃
So,
5250 g H₂ will produce = X g of NH₃

Solving for X,
X = (5250 g × 34 g) ÷ 6 g

X = 29750 g of NH₃
Or,
X = 29.75 Kg of NH₃