Nitric acid is produced commercially by the Ostwald process, represented by the following equations: $$ \begin{aligned} 4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ 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 given in the exercise: \[ \begin{aligned} 4 \mathrm{NH}_{3(g)} + 5 \mathrm{O}_{2(g)} \rightarrow 4 \mathrm{NO(g)} + 6 \mathrm{H}_{2}\mathrm{O(g)} \\ 2 \mathrm{NO(g)} + \mathrm{O}_{2(g)} \rightarrow 2 \mathrm{NO}_{2(g)} \\ 3 \mathrm{NO}_{2(g)} + \mathrm{H}_{2}\mathrm{O(l)} \rightarrow 2 \mathrm{HNO}_{3(aq)} + \mathrm{NO(g)} \end{aligned} \]

To find the moles of HNO3 produced, we'll use the equation: \(moles\ of\ HNO3 = \frac{mass\ of\ HNO3}{molar\ mass\ of\ HNO3}\) The molar mass of HNO3 (62.0 g/mol) can be determined from the periodic table by adding the molar masses of hydrogen, nitrogen, and oxygen, which are 1.0 g/mol, 14.0 g/mol, and 16.0 g/mol, respectively. The mass of HNO3 produced is given as 1.0 x 10^6 kg, which is equivalent to 1.0 x 10^9 g. Now we can find the moles of HNO3 produced: \(moles\ of\ HNO3 = \frac{1.0 \times 10^9\ g}{62.0\ g/mol} = 1.61 \times 10^7\ mol\)

To find the moles of NH3 required to produce the given amount of HNO3, we have to look at the stoichiometry of the overall reaction. For that, we need to add the three equations together: \[ 4 \mathrm{NH}_3 + 5 \mathrm{O}_2 + 2 \mathrm{NO} + \mathrm{O}_2 + 3 \mathrm{NO}_2 + \mathrm{H}_2\mathrm{O} \rightarrow 4 \mathrm{NO} + 6 \mathrm{H}_2\mathrm{O} + 2 \mathrm{NO}_2 + 2 \mathrm{HNO}_3 + \mathrm{NO} \] Canceling the common species on both sides of the equation, we get the net equation: \[ 4 \mathrm{NH}_3 + 9 \mathrm{H}_2\mathrm{O} \rightarrow 6 \mathrm{H}_2\mathrm{O} + 2 \mathrm{HNO}_3 \] Now, let's focus on the stoichiometric coefficients for NH3 and HNO3: \(4\ NH3 \rightarrow 2\ HNO3\) To find the moles of NH3 required to produce the given moles of HNO3, we can set up the following proportion: \(moles\ of\ NH3 = \frac{moles\ of\ HNO3 \times 4}{2}\) Now, we can plug in the moles of HNO3 we found in step 2: \(moles\ of\ NH3 = \frac{1.61 \times 10^7\ mol \times 4}{2} = 3.23 \times 10^7\ mol\)

To find the mass of NH3 required, we'll use the equation: \(mass\ of\ NH3 = moles\ of\ NH3 \times molar\ mass\ of\ NH3\) Using the periodic table, we can determine the molar mass of NH3 (17.0 g/mol) by adding the molar masses of nitrogen and hydrogen. Finally, we can find the mass of NH3 required: \(mass\ of\ NH3 = 3.23 \times 10^7\ mol\ NH3 \times 17.0 \frac{g}{mol\ NH3} = 5.49 \times 10^8\ g\) The required mass of NH3 to produce 1.0 x 10^6 kg HNO3 using the Ostwald process is \(5.49 \times 10^8\) g or \(5.49 \times 10^5\) kg, assuming a 100% yield and no recycling of NO.