`50.0 kg` of `N_(2) (g)` and `10 kg` of `H_(2) (g)` are mixed to produce `NH_(3) (g)`. Calculate the `NH_(3) (g)` formed. Identify the limiting reagent.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Convert the mass of \( N_2 \) and \( H_2 \) from kg to grams.
- Given mass of \( N_2 = 50.0 \, \text{kg} = 50.0 \times 1000 \, \text{g} = 50000 \, \text{g} \)
- Given mass of \( H_2 = 10.0 \, \text{kg} = 10.0 \times 1000 \, \text{g} = 10000 \, \text{g} \)

### Step 2: Calculate the number of moles of \( N_2 \) and \( H_2 \).
- Molecular weight of \( N_2 = 28 \, \text{g/mol} \) (since \( N = 14 \, \text{g/mol} \) and \( N_2 \) has 2 nitrogen atoms)
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