The following set of reactions is to be used as the basis of a method for producing nitric acid, \(\mathrm{HNO}_{3}\) Calculate the minimum masses of \(\mathrm{N}_{2}, \mathrm{H}_{2^{\prime}}\) and \(\mathrm{O}_{2}\) required per kilogram of \(\mathrm{HNO}_{3}\) $$\begin{array}{l} \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g}) \\ 4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g}) \\ 3 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow 2 \mathrm{HNO}_{3}(\mathrm{aq})+\mathrm{NO}(\mathrm{g}) \end{array}$$

Recall the molar masses of the atoms: \(\mathrm{N} = 14.007 \, \mathrm{gmol}^{-1}\), \(\mathrm{H} = 1.008 \, \mathrm{gmol}^{-1}\), \(\mathrm{O} = 16.00 \, \mathrm{gmol}^{-1}\). This will be needed to convert between mass and moles.

Calculate the moles of Nitric acid (\(\mathrm{HNO}_{3}\)) required. For Nitric acid, \(1 \, \mathrm{HNO}_{3}\) molecule has \(1 \, \mathrm{H}\), \(1 \, \mathrm{N}\) and \(3 \, \mathrm{O}\) atom(s). So, the molar mass of \(1 \, \mathrm{HNO}_{3}\) molecule is \(1.008 + 14.007 + 3 \times 16.00 = 63.012 \, \mathrm{gmol}^{-1}\). Thus, for 1 kilogram (or 1000 grams) of \(\mathrm{HNO}_{3}\), moles of \(\mathrm{HNO}_{3}\) required are \(1000 \, \mathrm{g} / \, 63.012 \, \mathrm{gmol}^{-1} = 15.873 \, \mathrm{mol}\).

Using stoichiometry of the reactions, calculate the required moles of \(\mathrm{N}_{2}\), \(\mathrm{H}_{2}\), and \(\mathrm{O}_{2}\). Looking at the reaction series, it can be seen that: for 2 moles of \(\mathrm{HNO}_{3}\) being produced, 3 moles of \(\mathrm{NO}_{2}\) are required. For 2 moles of \(\mathrm{NO}_{2}\), 2 moles of \(\mathrm{NO}\) are required. For 4 moles of \(\mathrm{NO}\), 4 moles of \(\mathrm{NH}_{3}\) are required. For 2 moles of \(\mathrm{NH}_{3}\), 1 mole of \(\mathrm{N}_{2}\) is required. For the Hydrogen and Oxygen, looking at all equations it shows that for 2 moles of \(\mathrm{HNO}_{3}\), 6 moles of \(\mathrm{H}_{2}\) and 11 moles of \(\mathrm{O}_{2}\) are required. Therefore the moles of \(\mathrm{N}_{2}\), \(\mathrm{H}_{2}\), and \(\mathrm{O}_{2}\) required to produce \(15.873 \, \mathrm{mol}\) of \(\mathrm{HNO}_{3}\) are: \(15.873/2 = 7.937 \, \mathrm{mol} \, \mathrm{N}_{2}\), \(15.873 \times 3 = 47.619 \, \mathrm{mol} \, \mathrm{H}_{2}\) and \(15.873 \times 11/2 = 87.415 \, \mathrm{mol} \, \mathrm{O}_{2}\).

Finally, convert the moles to grams. For nitrogen, mass is \(7.937 \, \mathrm{mol} \times 28.014 \, \mathrm{gmol}^{-1} = 222.41 \, \mathrm{g}\). For Hydrogen, mass is \(47.619 \, \mathrm{mol} \times 2.016 \, \mathrm{gmol}^{-1} = 95.93 \, \mathrm{g}\). For Oxygen, mass is \(87.415 \, \mathrm{mol} \times 32.00 \, \mathrm{gmol}^{-1} = 2797.28 \, \mathrm{g}\).