Calculation of NH3 formed and identification of limiting reagent:

Calculation of moles of N2 and H2:
- Moles of N2 = 50.0 kg / 28.0 g/mol = 1785.7 mol
- Moles of H2 = 10.0 kg / 2.0 g/mol = 5000 mol

Determination of limiting reagent:
- The balanced chemical equation for the production of NH3 is N2 + 3H2 → 2NH3
- The stoichiometric ratio of N2 to H2 is 1:3
- Therefore, 1 mol of N2 requires 3 mol of H2 to produce 2 mol of NH3
- In this case, the available moles of H2 (5000 mol) is more than enough to react with the moles of N2 (1785.7 mol) which means that N2 is the limiting reagent.

Calculation of NH3 formed:
- From the balanced chemical equation, 1 mol of N2 reacts to form 2 mol of NH3
- Therefore, 1785.7 mol of N2 will produce 3571.4 mol of NH3
- Mass of NH3 formed = 3571.4 mol x 17.0 g/mol = 60639.8 g or 60.6 kg

Explanation:

The given question involves the production of NH3 from N2 and H2. The balanced chemical equation for this reaction is N2 + 3H2 → 2NH3. The first step in solving this question is to calculate the moles of N2 and H2. This is done using the given masses and the molar masses of N2 and H2.

Next, we need to determine which of the two reactants is the limiting reagent. The limiting reagent is the reactant that is completely consumed in the reaction and determines the amount of product that can be formed. To find the limiting reagent, we use the stoichiometric ratios given by the balanced chemical equation.

In this case, we find that N2 is the limiting reagent because the available moles of H2 is more than enough to react with all the moles of N2. Once we have identified the limiting reagent, we can use the stoichiometric ratios to calculate the amount of product formed. In this case, we use the moles of N2 and the stoichiometric ratio (1 mol of N2 produces 2 mol of NH3) to find the moles of NH3 formed.

Finally, we convert the moles of NH3 to mass using the molar mass of NH3. The answer is 60.6 kg of NH3 formed.

NH3-56.67 kg