The mass of ammonia in grams produced when 2.8 kg of dinitrogen quantitatively reacts with 1 kg of dihydrogen is __________

We begin with the balanced chemical equation for the synthesis of ammonia:

$$\mathrm{N_2 + 3\,H_2 \;\longrightarrow\; 2\,NH_3}$$

From this equation we note the stoichiometric mole ratio:

$$\mathrm{N_2 : H_2 : NH_3 \;=\; 1 : 3 : 2}$$

Now we convert each given mass into moles because stoichiometric calculations are always performed in terms of moles.

The molar mass of dinitrogen is $$\mathrm{28\;g\,mol^{-1}}$$, so for $$2.8\;\text{kg}$$ (which equals $$2800\;\text{g}$$) the number of moles of dinitrogen is

$$n_{\mathrm{N_2}}=\dfrac{2800\;\text{g}}{28\;\text{g\,mol}^{-1}}=100\;\text{mol}$$

The molar mass of dihydrogen is $$\mathrm{2\;g\,mol^{-1}}$$, so for $$1\;\text{kg}$$ (which equals $$1000\;\text{g}$$) the number of moles of dihydrogen is

$$n_{\mathrm{H_2}}=\dfrac{1000\;\text{g}}{2\;\text{g\,mol}^{-1}}=500\;\text{mol}$$

Next we identify the limiting reagent by comparing the available mole ratio with the required stoichiometric ratio.

According to the balanced equation, $$1$$ mole of $$\mathrm{N_2}$$ needs $$3$$ moles of $$\mathrm{H_2}$$. Therefore, for the $$100$$ moles of $$\mathrm{N_2}$$ present, the hydrogen required would be

$$100 \times 3 = 300\;\text{mol of } \mathrm{H_2}$$

We actually have $$500\;\text{mol of } \mathrm{H_2}$$, which is more than the $$300\;\text{mol}$$ required. Hence hydrogen is present in excess, and nitrogen is the limiting reagent.

Since $$\mathrm{N_2}$$ is the limiting reagent, the amount of ammonia formed depends entirely on the moles of $$\mathrm{N_2}$$ available. The balanced equation tells us that $$1$$ mole of $$\mathrm{N_2}$$ produces $$2$$ moles of $$\mathrm{NH_3}$$. Therefore, for the $$100$$ moles of $$\mathrm{N_2}$$ the moles of ammonia produced will be

$$n_{\mathrm{NH_3}} = 100 \times 2 = 200\;\text{mol}$$

Finally, we convert these moles of ammonia into mass. The molar mass of ammonia is $$\mathrm{17\;g\,mol^{-1}}$$, so

$$m_{\mathrm{NH_3}} = 200\;\text{mol} \times 17\;\text{g\,mol}^{-1} = 3400\;\text{g}$$

This value can also be expressed as $$3.4\;\text{kg}$$, but since the question asks for grams, we keep it as $$3400\;\text{g}$$.

So, the answer is $$3400 \text{ g}$$.