Question

Industrially, nitric acid is produced by the Ostwald process represented by the following equations: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow & 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 2 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{HNO}_{3}(a q)+\mathrm{HNO}_{2}(a q) \end{aligned} $$ What mass of \(\mathrm{NH}_{3}\) (in grams) must be used to produce 1.00 ton of \(\mathrm{HNO}_{3}\) by the Ostwald process, assuming an 80 percent yield in each step \((1\) ton \(=2000 \mathrm{lb} ;\) $$ 1 \mathrm{lb}=453.6 \mathrm{~g}) ? $$

Short answer

383,400 grams of \(\mathrm{NH}_{3}\) are required.

Step-by-step solution

Convert tons to grams

First, let's convert the mass of nitric acid from tons to grams. We have 1.00 ton of \(\mathrm{HNO}_{3}\), so we use the conversion factors: \[ 1 \text{ ton} = 2000 \text{ lb} \] and \[ 1 \text{ lb} = 453.6 \text{ g} \]. Calculate the mass in grams: \[ 1 \text{ ton} \times 2000 \text{ lb/ton} \times 453.6 \text{ g/lb} = 907200 \text{ g} \].

Find moles of \(\mathrm{HNO}_{3}\)

Next, we determine the number of moles of \(\mathrm{HNO}_{3}\) in 907200 grams. The molar mass of \(\mathrm{HNO}_{3}\) is: \[ M = 1\times14.01\,(\mathrm{N}) + 1\times1.01\,(\mathrm{H}) + 3\times16.00\,(\mathrm{O}) = 63.02 \text{ g/mol} \]. Number of moles of \(\mathrm{HNO}_{3}\): \[ \text{Moles} = \frac{907200 \text{ g}}{63.02 \text{ g/mol}} \approx 14400 \text{ mol} \].

Calculate required \(\mathrm{NO}\) moles

From the third reaction equation, \[ 2 \mathrm{NO}_{2} + \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{HNO}_{3} + \mathrm{HNO}_{2} \], we see that 1 mole of \(\mathrm{NO}_{2}\) forms 1 mole of \(\mathrm{HNO}_{3}\). Thus, approximately 14400 moles of \(\mathrm{NO}_{2}\) are needed. Each mole of \(\mathrm{NO}_{2}\) requires 1 mole of \(\mathrm{NO}\) from the second reaction equation: \[ 2 \mathrm{NO} + \mathrm{O}_{2} \rightarrow 2 \mathrm{NO}_{2} \]. Therefore, 14400 moles of \(\mathrm{NO}\) are also needed.

Account for 80% yield per step

Since each step of the process yields 80% of the expected product, we account for this efficiency through the entire process. We figure out the initial moles of \(\mathrm{NO}\) that must be produced at 100% efficiency to result in 14400 moles accounting for yields:\[ \text{Initial moles of } \mathrm{NO} = \frac{14400}{(0.8)^{2}} \approx 22500 \]. This compensates for the 20% loss in each step leading up to the formation of \(\mathrm{HNO}_{3}\).

Calculate required \(\mathrm{NH}_{3}\) moles

Using the stoichiometry from the first reaction,\[ 4 \mathrm{NH}_{3} + 5 \mathrm{O}_{2} \rightarrow 4 \mathrm{NO} + 6 \mathrm{H}_{2} \mathrm{O} \], we see that 4 moles of \(\mathrm{NH}_{3}\) produce 4 moles of \(\mathrm{NO}\). Therefore, 22500 moles of \(\mathrm{NO}\) require 22500 moles of \(\mathrm{NH}_{3}\).

Calculate mass of \(\mathrm{NH}_{3}\)

Now calculate the mass of \(\mathrm{NH}_{3}\) required. The molar mass of \(\mathrm{NH}_{3}\) is: \[ 1\times14.01\,(\mathrm{N}) + 3\times1.01\,(\mathrm{H}) = 17.04 \text{ g/mol} \]. Thus, the mass is: \[ \text{Mass} = 22500 \text{ mol} \times 17.04 \text{ g/mol} = 383400 \text{ g} \].

Key concepts

Nitric Acid Production

The production of nitric acid is a fundamental process in the chemical industry. It primarily uses the Ostwald process, which is highly effective for large-scale manufacturing. Nitric acid (\(\text{HNO}_3\)) is essential in producing fertilizers, explosives, and dyes. Understanding the Ostwald process helps us grasp how industrial-scale chemical reactions function.
This process consists of three reactions:

• Ammonia oxidation: \(4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightarrow 4 \text{NO}(g) + 6 \text{H}_2\text{O}(l)\)
• \(2 \text{NO}(g) + \text{O}_2(g) \rightarrow 2 \text{NO}_2(g)\)
• \(2 \text{NO}_2(g) + \text{H}_2 \text{O}(l) \rightarrow \text{HNO}_3(aq) + \text{HNO}_2(aq)\)

The process efficiently converts ammonia into nitric acid through oxidation and absorption, essential for meeting industrial demands.

Chemical Stoichiometry

Chemical stoichiometry involves the calculation of reactants and products in chemical reactions. It ensures precise ratios of materials are used, preventing waste and optimizing outcomes.
The stoichiometric calculations in the Ostwald process determine how much ammonia (\(\text{NH}_3\)) is necessary to yield desired nitric acid quantities. By analyzing each reaction, stoichiometry helps us track reactant quantities and predict product amounts. For instance, the reaction \(4 \text{NH}_3 + 5 \text{O}_2 \rightarrow 4 \text{NO} + 6 \text{H}_2 \text{O}\)
requires a precise 4:5 ratio of ammonia to oxygen. Understanding these ratios is crucial for accurate chemical engineering and ensuring each stage of production proceeds efficiently.

Ammonia Oxidation

Ammonia oxidation is a key step in the Ostwald process. It involves the conversion of ammonia (\(\text{NH}_3\)) to nitric oxide (\(\text{NO}\)), which is then further oxidized to nitrogen dioxide (\(\text{NO}_2\)). This initial oxidation is represented by:

• \(4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightarrow 4 \text{NO}(g) + 6 \text{H}_2 \text{O}(l)\)

This reaction occurs at high temperature and pressure in the presence of a platinum catalyst. It is a crucial step because it forms the intermediate compounds necessary for producing nitric acid. Efficient ammonia oxidation is vital for minimizing resource use and maximizing output. Monitoring conditions like temperature and pressure ensures optimal performance and high yields.

Industrial Chemistry

Industrial chemistry focuses on the large-scale production of chemicals, applying principles of chemistry and engineering to meet the demands of society. It involves processes like the Ostwald process, which exemplifies how industrial reactions are scaled up from lab techniques.
The Ostwald process showcases the intricacies of managing reactions over large volumes. Factors such as reaction kinetics, catalyst efficiency, and heat management are critical to ensuring that reactions are safe and economically viable. Industrial chemistry also considers environmental impacts, implementing methods to reduce emissions and waste. By balancing these elements, industrial chemistry plays a significant role in sustaining modern infrastructure and economy.