Nitric acid is produced commercially by the Ostwald process, represented by the following equations: $$\begin{array}{c}4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\\2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \\\3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g)\end{array}$$ 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.

We know that the mass of HNO₃ (m₍ₕₙᵒ₃₎) we want to produce is 1.0 x 10⁶ kg. We can convert this to grams (1 kg = 1000 g) and then use the molar mass of HNO₃ (M₍ₕₙᵒ₃₎ = 63 g/mol) to find the moles of HNO₃ required. Moles of HNO₃ (n₍ₕₙᵒ₃₎) = (m₍ₕₙᵒ₃₎ x 1000) / M₍ₕₙᵒ₃₎

Now, we're going to use stoichiometry with the balanced equations to determine the moles of NH₃ required to produce the desired amount of HNO₃. From the third equation, we see that 3 moles of NO₂ produce 2 moles of HNO₃. We can express this relationship using a ratio between moles of NO₂ (n₍ₙᵒ₂₎) and moles of HNO₃ (n₍ₕₙᵒ₃₎): n₍ₙᵒ₂₎ = (3/2) × n₍ₕₙᵒ₃₎ Next, from the second equation, we can find out how many moles of NO are required to produce the moles of NO₂ calculated above: n₍ₙᵒ₂₎ = 2 × n_NO Now, let's find the moles of NH₃ (n₍ₙₕ₃₎) required to produce the amount of moles of NO: From the first equation, we see that 4 moles of NH₃ produce 4 moles of NO. We can express this relationship using a ratio between moles of NH₃ (n₍ₙₕ₃₎) and moles of NO (n_NO) as follows: n₍ₙₕ₃₎ = n_NO Putting everything together, we have: n₍ₙₕ₃₎ = n_NO = (1/2) × n₍ₙᵒ₂₎ = (1/2) × (3/2) × n₍ₕₙᵒ₃₎ Now, we can plug in the moles of HNO₃ calculated in Step 1 to find the moles of NH₃ required.

We have found the moles of NH₃ required, now we can convert it to mass. Firstly, we will find the molar mass of NH₃ (M₍ₙₕ₃₎ = 17 g/mol). Mass of NH₃ (m₍ₙₕ₃₎) = n₍ₙₕ₃₎ × M₍ₙₕ₃₎ Using all this information, we can find the mass of NH₃ required to produce 1.0 x 10⁶ kg of HNO₃ by the Ostwald process.