Nitric acid is produced commercially by the Ostwald process, represented by the following equations: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ What mass of \(\mathrm{NH}_{3}\) must be used to produce $1.0 \times 10^{6} \mathrm{kg}\( \)\mathrm{HNO}_{3}\( by the Ostwald process? Assume 100\)\%$ yield in each reaction, and assume that the NO produced in the third step is not recycled.

The balanced chemical equations for the Ostwald process are: \(4 NH_{3}(g) + 5 O_{2}(g) \longrightarrow 4 NO(g) + 6 H_{2}O(g)\) \(2 NO(g) + O_{2}(g) \longrightarrow 2 NO_{2}(g)\) \(3 NO_{2}(g) + H_{2}O(l) \longrightarrow 2 HNO_{3}(aq) + NO(g)\)

From the balanced equations, we can see the following stoichiometric relationships: For every 4 moles of NH3, we can produce 4 moles of NO in Reaction 1. For every 4 moles of NO, we can produce 4 moles of NO2 in Reaction 2. For every 6 moles of NO2, we can produce 4 moles of HNO3 in Reaction 3. To produce 1 mole of HNO3, we need: \[\frac{1}{4} \times 6 = 1.5\] moles of NO2 (from the last equation) By stoichiometry, we can find the amount of NO that leads to 1.5 moles of NO2: \[\frac{1.5 \text{ moles of } NO2}{2} =0.75 \text{ moles of } NO \text{ (from the second equation)}\] Now, we can find how many moles of NH3 leads to 0.75 moles of NO: \[\frac{0.75 \text{ moles of } NO}{4} = 0.1875 \text{ moles of } NH3\text{ (from the first equation)}\]

We are asked to produce \(1.0 \times 10^{6} kg\) of HNO3. First, we will convert the mass to moles using the molar mass of HNO3: Molar mass of HNO3 = \(1 \times 1.008 + 1 \times 14.01 + 3 \times 16\) = 63.02 g/mol Amount of HNO3 required (in moles) = \(\frac{1.0 \times 10^{6} kg}{63.02 g/mol}\) = \(1.5875 \times 10^{7} mol\)

We have determined that 1 mole of HNO3 needs 0.1875 moles of NH3. Therefore, to produce \(1.5875 \times 10^{7} mol\) of HNO3, we need: Amount of NH3 required (in moles) = \(0.1875 \times 1.5875 \times 10^{7} mol = 2.9781 \times 10^{6} mol \)

Finally, we'll convert the moles of NH3 to mass using its molar mass: Molar mass of NH3 = \(1 \times 14.01 + 3 \times 1.008\) = 17.031 g/mol Mass of NH3 required = \((2.9781 \times 10^{6} mol)(17.031 g/mol) = 5.0748 \times 10^{7} g\) Converting to kg: \(5.0748 \times 10^{7} g = 5.0748 \times 10^{4} kg\) Thus, 50,748 kg of NH3 must be used to produce \(1.0 \times 10^{6} kg\) of HNO3 by the Ostwald process, assuming a 100% yield in each reaction and no recycling of NO.