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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 g) must be used to produce 1.00 ton of \(\mathrm{HNO}_{3}\) by the above procedure, assuming an 80 percent yield in each step? \((1\) ton \(=2000 \mathrm{lb}\); \(1 \mathrm{lb}=453.6 \mathrm{~g} .)\)

Short Answer

Expert verified
To produce one ton of \(HNO_{3}\) by the given procedure, approximately \(381200\) grams (or 381.2kg) \(NH_{3}\) are needed.

Step by step solution

01

Convert ton to grams

We need to convert the desired mass of \(HNO_{3}\) from tons to grams, using the provided equivalences. Doing the conversion, we find that 1 ton is equal to \(1 \times 2000 \times 453.6 = 907200\) grams.
02

Stoichiometry from balanced chemical equations

From the balanced equations, it can be found that every 2 moles of \(NO_{2}\) produces 1 mole of \(HNO_{3}\), every 2 moles of \(NO\) is converted into 2 moles of \(NO_{2}\), and 4 moles of \(NH_{3}\) reacts to produce 4 moles of \(NO\). Therefore, we have a 1:1:1:2 ratio of \(HNO_{3}\): \(NO_{2}\): \(NO\): \(NH_{3}\). This shows that one mole of \(NH_{3}\) is required to produce one mole of \(HNO_{3}\).
03

Calculate molar mass

By adding up the atomic masses from the periodic table, we find the molar mass of \(NH_{3}\) to be approximately 17 g/mol, and the molar mass of \(HNO_{3}\) to be approximately 63 g/mol.
04

Adjust for reaction yield

With 80% efficiency in each step, we have to account for this yield in our calculations. So, for each gram of \(HNO_{3}\) we wish to produce, \(1 ÷ 0.8 = 1.25\) grams of \(NH_{3}\) is required.
05

Final Calculation

Combining all this information and considering the yield factor, we calculate the required \(NH_{3}\) mass for the production of 1 ton of \(HNO_{3}\). That is \(907200 (g of HNO_{3}) \times 1.25 (quite an efficiency) \times 17/63 (conversion factor from g of HNO_{3} to g of NH_{3})\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equation Balancing
Understanding how to balance chemical equations is crucial in chemistry, especially when dealing with reactions like the Ostwald process for the production of nitric acid.

Balancing chemical equations involves ensuring that the number of atoms for each element is equal on both the reactant and product sides of the equation. This is essential due to the conservation of mass in chemical reactions; matter cannot be created or destroyed.

In the Ostwald process, the sequence of reactions starts with ammonia (\textbf{NH}\(_3\)) and oxygen (\textbf{O}\(_2\)) reacting to form nitrogen monoxide (\textbf{NO}) and water (\textbf{H}\(_2\)\textbf{O}). This is followed by the oxidation of \textbf{NO} to nitrogen dioxide (\textbf{NO}\(_2\)), which then reacts with water to produce nitric acid (\textbf{HNO}\(_3\)) and nitrous acid (\textbf{HNO}\(_2\)). Each step must be balanced to reflect the conservation of atoms:
  • 4 \textbf{NH}\(_3\)(g) + 5 \textbf{O}\(_2\)(g) → 4 \textbf{NO}(g) + 6 \textbf{H}\(_2\) \textbf{O}(l)
  • 2 \textbf{NO}(g) + \textbf{O}\(_2\)(g) → 2 \textbf{NO}\(_2\)(g)
  • 3 \textbf{NO}\(_2\)(g) + \textbf{H}\(_2\) \textbf{O}(l) → \textbf{HNO}\(_3\)(aq) + \textbf{HNO}\(_2\)(aq)

Such balanced equations are critical as they serve as a map for stoichiometric calculations, allowing us to quantitatively describe the relationships between reactants and products.
Stoichiometry
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. Using stoichiometry, one can predict how much reactant is needed to produce a desired amount of product, a concept applied in industrial processes.

The stoichiometry of the Ostwald process demonstrates these relationships explicitly. In the production of nitric acid, each molecule of ammonia contributes to the creation of a molecule of nitric acid, in principle. In our exercise, these relationships are defined by the molar ratios derived from the chemical equations.

It is critical to convert all quantities to moles by using their molar masses to understand the actual amounts involved. For example, to know how much ammonia is needed to produce a certain mass of nitric acid, one must determine the molar mass of both compounds and use the molar ratio obtained from the balanced chemical equations.

However, this theoretical amount is adjusted for real world applications due to yields being less than 100% efficient. In this instance, we have an 80% yield at each step, which alters the straightforward stoichiometric calculation, making it necessary to increase the amount of ammonia to compensate for the loss in efficiency.
Industrial Production of Nitric Acid
The industrial production of nitric acid through the Ostwald process is a significant application of chemical engineering principles, combining concepts of stoichiometry and chemical equation balancing into a large-scale production methodology.

This process includes a series of steps beginning with the catalytic oxidation of ammonia. After ammonia is oxidized to produce nitrogen monoxide, it further reacts to form nitrogen dioxide. Finally, the nitrogen dioxide reacts with water to yield nitric acid, which is highly desirable for a variety of uses in agriculture, manufacturing, and even in the production of explosives.

Efficiency plays a pivotal role in industrial processes. Idealized chemical reactions do not always align with practical outcomes, due to factors like incomplete reactions, side reactions, and loss of materials. As such, the yield from each step of a process like the Ostwald process must be considered when determining the amount of starting materials required to achieve a desired quantity of product.

In the case of the exercise, converting from theoretical stoichiometry to practical application necessitates adjusting the amount of ammonia needed by accounting for the 80% yield at each reaction step, reflecting a realistic industrial scenario. Industries use similar calculations to scale their production quantities while minimizing waste and cost.

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Most popular questions from this chapter

Why must a chemical equation be balanced? What law is obeyed by a balanced chemical equation?

(a) A research chemist used a mass spectrometer to study the two isotopes of an element. Over time, she recorded a number of mass spectra of these isotopes. On analysis, she noticed that the ratio of the taller peak (the more abundant isotope) to the shorter peak (the less abundant isotope) gradually increased with time. Assuming that the mass spectrometer was functioning normally, what do you think was causing this change? (b) Mass spectrometry can be used to identify the formulas of molecules having small molecular masses. To illustrate this point, identify the molecule which most likely accounts for the observation of a peak in a mass spectrum at: 16 amu, 17 amu, 18 amu, and 64 amu. (c) Note that there are (among others) two likely molecules that would give rise to a peak at 44 amu, namely, \(\mathrm{C}_{3} \mathrm{H}_{8}\) and \(\mathrm{CO}_{2}\). In such cases, a chemist might try to look for other peaks generated when some of the molecules break apart in the spectrometer. For example, if a chemist sees a peak at 44 amu and also one at 15 amu, which molecule is producing the 44 -amu peak? Why? (d) Using the following precise atomic masses: \({ }^{1} \mathrm{H}(1.00797\) amu \(),{ }^{12} \mathrm{C}(12.00000 \mathrm{amu}),\) and \({ }^{16} \mathrm{O}(15.99491 \mathrm{amu}),\) how precisely must the masses of \(\mathrm{C}_{3} \mathrm{H}_{8}\) and \(\mathrm{CO}_{2}\) be measured to distinguish between them? (e) Every year millions of dollars' worth of gold is stolen. In most cases the gold is melted down and shipped abroad. This way the gold retains its value while losing all means of identification. Gold is a highly unreactive metal that exists in nature in the uncombined form. During the mineralization of gold, that is, the formation of gold nuggets from microscopic gold particles, various elements such as cadmium (Cd), lead (Pb), and zinc \((\mathrm{Zn})\) are incorporated into the nuggets. The amounts and types of the impurities or trace elements in gold vary according to the location where it was mined. Based on this knowledge, describe how you would identify the source of a piece of gold suspected of being stolen from Fort Knox, the federal gold depository.

Tin(II) fluoride \(\left(\mathrm{SnF}_{2}\right)\) is often added to toothpaste as an ingredient to prevent tooth decay. What is the mass of \(\mathrm{F}\) in grams in \(24.6 \mathrm{~g}\) of the compound?

A mixture of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\) is heated until all the water is lost. If \(5.020 \mathrm{~g}\) of the mixture gives \(2.988 \mathrm{~g}\) of the anhydrous salts, what is the percent by mass of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) in the mixture?

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