What is the theoretical yield of ammonia, in kilograms, that we can synthesize from 5.22 kg of h2 and 31.5 kg of n2?

The calculated theoretical yield of ammonia is 29.58 Kg

In the solution provided,

the mass of is 5.22 Kg, which equals 5220 g

the mass of is 31.5 Kg, equivalent to 31500 g

The molar mass of is 2 g/mole

The molar mass of is 28 g/mole

The molar mass of is 17 g/mole

First, we must calculate the moles of and .

Moles of =

Moles of =

Next, we need to find which reagent is limiting and which is excess.

The balanced equation for the reaction will be,

Based on the balanced equation, we find that

1 mole of reacts with 3 moles of

1125 moles of react with moles of

This indicates that is the limiting reagent and is in excess.

Now, we proceed to calculate the moles of ammonia.

According to the reaction, we determine that,

3 moles of will yield 2 moles of ammonia

2610 moles of will provide moles of ammonia

Finally, we will calculate the mass of the ammonia produced.

Therefore, the theoretical yield of ammonia is 29.58 Kg

The following data is provided:

The mass of H2 is 5.22 kg, which is equal to 5220 g

The mass of N2 is 31.5 kg, or 31500 g

We need to discover:

Theoretical yield of NH3

The explanation is as follows:

The balanced chemical equation is:

N2 + 3H2 → 2NH3

1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3

Calculating moles of N2, we get: # moles of N2 = 31500 g/ 28 g.mol-1 = 1125 moles

Calculating moles of H2, we find: # moles of H2 = 5200 g / 1 g.mol-1 = 5200 moles

Therefore, N2 acts as the limiting reagent

According to stoichiometry:

1 mole of N2 creates 2 moles of NH3

Thus, with 1125 moles of N2, we yield: 1125 * 2 = 2250 moles of NH3

Calculating the mass of NH3: Mass of NH3 = 2250 moles * 17 g/mole = 38250 g = 38.3 kg

Thus, the theoretical yield of NH3 equals 38.3 kg