Given:
- Mass of N2(g) = 50.0 kg
- Mass of H2(g) = 10.0 kg

To find:
- Number of moles of NH3(g) formed
- Limiting reagent in the production of NH3

Solution:

Step 1: Calculate Moles of N2 and H2
To calculate the number of moles, we will use the formula:

Number of moles = Mass / Molar mass

The molar mass of N2 is 28.0134 g/mol, and for H2, it is 2.01588 g/mol.

Number of moles of N2 = 50.0 kg / 28.0134 g/mol
Number of moles of H2 = 10.0 kg / 2.01588 g/mol

Step 2: Determine the Stoichiometry of the Reaction
The balanced chemical equation for the production of NH3 from N2 and H2 is:

N2(g) + 3H2(g) → 2NH3(g)

According to the stoichiometry of the reaction, 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.

Step 3: Determine the Limiting Reagent
The limiting reagent is the reactant that is completely consumed in the reaction, thereby limiting the amount of product formed. To identify the limiting reagent, we compare the moles of each reactant with the stoichiometry of the reaction.

For N2:
Number of moles of N2 = 50.0 kg / 28.0134 g/mol = 1784.5 mol
According to the stoichiometry, 1 mole of N2 reacts with 2 moles of NH3.
Therefore, the moles of NH3 that can be formed from N2 = 1784.5 mol * (2 mol NH3 / 1 mol N2) = 3569 mol

For H2:
Number of moles of H2 = 10.0 kg / 2.01588 g/mol = 496.4 mol
According to the stoichiometry, 3 moles of H2 react with 2 moles of NH3.
Therefore, the moles of NH3 that can be formed from H2 = 496.4 mol * (2 mol NH3 / 3 mol H2) = 330.93 mol

Conclusion:
From the calculations, we can see that the moles of NH3 that can be formed from N2 is 3569 mol, while from H2 it is 330.93 mol. Therefore, the limiting reagent in this situation is H2 (hydrogen gas) because it limits the amount of NH3 produced.