1. Write the balanced chemical equation

$\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}$

2. Convert given masses into moles

• Mass of dinitrogen: $2.8\text{ kg} = 2800\text{ g}$

Molar mass of $\mathrm{N_2}$: $28\text{ g mol}^{-1}$

So, moles of $\mathrm{N_2}$: $\frac{2800}{28} = 100\text{ mol}$

• Mass of dihydrogen: $1\text{ kg} = 1000\text{ g}$

Molar mass of $\mathrm{H_2}$: $2\text{ g mol}^{-1}$

So, moles of $\mathrm{H_2}$: $\frac{1000}{2} = 500\text{ mol}$

3. Find the limiting reagent

From the equation, $1\text{ mol } \mathrm{N_2} \text{ requires } 3\text{ mol } \mathrm{H_2}$

For $100$ mol $\mathrm{N_2}$, required $\mathrm{H_2}$ is: $100 \times 3 = 300\text{ mol}$

Available $\mathrm{H_2}$ is $500$ mol, which is excess.

Therefore, $\mathrm{N_2}$ is the limiting reagent.

4. Calculate moles of ammonia formed

From the balanced equation, $1\text{ mol } \mathrm{N_2} \rightarrow 2\text{ mol } \mathrm{NH_3}$

So, $100\text{ mol } \mathrm{N_2} \rightarrow 200\text{ mol } \mathrm{NH_3}$

5. Convert moles of ammonia to mass

Molar mass of $\mathrm{NH_3}$: $14 + 3(1) = 17\text{ g mol}^{-1}$

Mass of $200$ mol $\mathrm{NH_3}$: $200 \times 17 = 3400\text{ g}$

6. Final answer

The mass of ammonia produced is: $\boxed{3400}$

gram.