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Chapter 4: Problem 90

The following set of reactions is to be used as the basis of a method for producing nitric acid, \(\mathrm{HNO}_{3}\) Calculate the minimum masses of \(\mathrm{N}_{2}, \mathrm{H}_{2^{\prime}}\) and \(\mathrm{O}_{2}\) required per kilogram of \(\mathrm{HNO}_{3}\) $$\begin{array}{l} \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g}) \\ 4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g}) \\ 3 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow 2 \mathrm{HNO}_{3}(\mathrm{aq})+\mathrm{NO}(\mathrm{g}) \end{array}$$

Short Answer

Expert verified
To produce 1 Kg of Nitric acid, minimum 222.41g of Nitrogen, 95.93g of Hydrogen and 2797.28g of Oxygen are needed.

Step by step solution

01

Recall Molar Mass

Recall the molar masses of the atoms: \(\mathrm{N} = 14.007 \, \mathrm{gmol}^{-1}\), \(\mathrm{H} = 1.008 \, \mathrm{gmol}^{-1}\), \(\mathrm{O} = 16.00 \, \mathrm{gmol}^{-1}\). This will be needed to convert between mass and moles.
02

Calculate Required Moles of HNO3

Calculate the moles of Nitric acid (\(\mathrm{HNO}_{3}\)) required. For Nitric acid, \(1 \, \mathrm{HNO}_{3}\) molecule has \(1 \, \mathrm{H}\), \(1 \, \mathrm{N}\) and \(3 \, \mathrm{O}\) atom(s). So, the molar mass of \(1 \, \mathrm{HNO}_{3}\) molecule is \(1.008 + 14.007 + 3 \times 16.00 = 63.012 \, \mathrm{gmol}^{-1}\). Thus, for 1 kilogram (or 1000 grams) of \(\mathrm{HNO}_{3}\), moles of \(\mathrm{HNO}_{3}\) required are \(1000 \, \mathrm{g} / \, 63.012 \, \mathrm{gmol}^{-1} = 15.873 \, \mathrm{mol}\).
03

Calculate Required Moles of N2, H2 and O2

Using stoichiometry of the reactions, calculate the required moles of \(\mathrm{N}_{2}\), \(\mathrm{H}_{2}\), and \(\mathrm{O}_{2}\). Looking at the reaction series, it can be seen that: for 2 moles of \(\mathrm{HNO}_{3}\) being produced, 3 moles of \(\mathrm{NO}_{2}\) are required. For 2 moles of \(\mathrm{NO}_{2}\), 2 moles of \(\mathrm{NO}\) are required. For 4 moles of \(\mathrm{NO}\), 4 moles of \(\mathrm{NH}_{3}\) are required. For 2 moles of \(\mathrm{NH}_{3}\), 1 mole of \(\mathrm{N}_{2}\) is required. For the Hydrogen and Oxygen, looking at all equations it shows that for 2 moles of \(\mathrm{HNO}_{3}\), 6 moles of \(\mathrm{H}_{2}\) and 11 moles of \(\mathrm{O}_{2}\) are required. Therefore the moles of \(\mathrm{N}_{2}\), \(\mathrm{H}_{2}\), and \(\mathrm{O}_{2}\) required to produce \(15.873 \, \mathrm{mol}\) of \(\mathrm{HNO}_{3}\) are: \(15.873/2 = 7.937 \, \mathrm{mol} \, \mathrm{N}_{2}\), \(15.873 \times 3 = 47.619 \, \mathrm{mol} \, \mathrm{H}_{2}\) and \(15.873 \times 11/2 = 87.415 \, \mathrm{mol} \, \mathrm{O}_{2}\).
04

Convert Moles to Grams

Finally, convert the moles to grams. For nitrogen, mass is \(7.937 \, \mathrm{mol} \times 28.014 \, \mathrm{gmol}^{-1} = 222.41 \, \mathrm{g}\). For Hydrogen, mass is \(47.619 \, \mathrm{mol} \times 2.016 \, \mathrm{gmol}^{-1} = 95.93 \, \mathrm{g}\). For Oxygen, mass is \(87.415 \, \mathrm{mol} \times 32.00 \, \mathrm{gmol}^{-1} = 2797.28 \, \mathrm{g}\).

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Key Concepts

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

Chemical Reactions
In the world of chemistry, a chemical reaction is a process where substances, known as reactants, are transformed into different substances, called products. This transformation involves the breaking of bonds in the reactants and the formation of new bonds to create the products.

In the exercise provided, there are several sequential reactions involved in the production of nitric acid (\(\mathrm{HNO}_{3}\)). Understanding these reactions is crucial to determining the amounts of starting materials needed:
  • First, nitrogen gas (\(\mathrm{N}_{2}\)) reacts with hydrogen gas (\(\mathrm{H}_{2}\)) to form ammonia (\(\mathrm{NH}_{3}\)).
  • Ammonia is then oxidized to form nitric oxide (\(\mathrm{NO}\)), which is further oxidized into nitrogen dioxide (\(\mathrm{NO}_{2}\)).
  • Finally, nitrogen dioxide reacts with water to form nitric acid (\(\mathrm{HNO}_{3}\)) and more nitric oxide.
These reactions are linked in a chain, meaning the products of one reaction become the reactants for the next. The stoichiometry, which is the quantitative relationship between reactants and products in these chemical reactions, is fundamental in calculating the required amounts of each substance.
Molar Mass Calculation
Molar mass is a key concept in chemistry that refers to the mass of one mole of a given substance. It is expressed in grams per mole (\(\text{g/mol}\)). Calculating the molar mass is crucial for converting between grams and moles, since chemists often need to measure quantities of substances in moles rather than grams.

In our exercise, knowing the molar mass of each atom is vital:
  • Nitrogen (\(\mathrm{N}\)): 14.007 g/mol
  • Hydrogen (\(\mathrm{H}\)): 1.008 g/mol
  • Oxygen (\(\mathrm{O}\)): 16.00 g/mol
For example, to find the molar mass of nitric acid (\(\mathrm{HNO}_{3}\)), add together the molar masses of its constituent atoms: 1 Hydrogen, 1 Nitrogen, and 3 Oxygen atoms, calculating as follows: \(1.008 + 14.007 + 3 imes 16.00 = 63.012\) g/mol.

This calculation allows us to convert the mass needed into moles, which directly relates to the stoichiometry of the reactions involved.
Nitric Acid Production
Nitric acid (\(\mathrm{HNO}_{3}\)) is an important industrial chemical with various applications, such as in fertilizers, explosives, and even in making rocket propellant. Understanding its production process on a molecular level helps us appreciate the underlying chemistry and the resources required to manufacture it efficiently.

In the set of reactions given, nitric acid is produced through a sequence of four reactions. Each step is essential in ensuring that nitrogen, hydrogen, and oxygen atoms end up in the correct composition necessary for producing nitric acid:
  • Initially, \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) combine to produce \(\mathrm{NH}_{3}\), a critical intermediate.
  • Next, \(\mathrm{NH}_{3}\) is oxidized by \(\mathrm{O}_{2}\) to form \(\mathrm{NO}\), which subsequently is oxidized to \(\mathrm{NO}_{2}\).
  • Finally, \(\mathrm{NO}_{2}\) undergoes a reaction with water, converting to \(\mathrm{HNO}_{3}\).
By understanding this full cycle, we can calculate precisely the amounts of \(\mathrm{N}_{2}\), \(\mathrm{H}_{2}\), and \(\mathrm{O}_{2}\) required to produce a desired quantity of \(\mathrm{HNO}_{3}\), as demonstrated in the original exercise.

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Most popular questions from this chapter

Lead nitrate and potassium iodide react in aqueous solution to form a yellow precipitate of lead iodide. In one series of experiments, the masses of the two reactants were varied, but the total mass of the two was held constant at \(5.000 \mathrm{g}\). The lead iodide formed was filtered from solution, washed, dried, and weighed. The table gives data for a series of reactions. $$\begin{array}{lll} \hline & \text { Mass of Lead } & \text { Mass of Lead } \\ \text { Experiment } & \text { Nitrate, } g & \text { lodide, } g \\ \hline 1 & 0.500 & 0.692 \\ 2 & 1.000 & 1.388 \\ 3 & 1.500 & 2.093 \\ 4 & 3.000 & 2.778 \\ 5 & 4.000 & 1.391 \\ \hline \end{array}$$ (a) Plot the data in a graph of mass of lead iodide versus mass of lead nitrate, and draw the appropriate curve(s) connecting the data points. What is the maximum mass of precipitate that can be obtained? (b) Explain why the maximum mass of precipitate is obtained when the reactants are in their stoichiometric proportions. What are these stoichiometric proportions expressed as a mass ratio, and as a mole ratio? (c) Show how the stoichiometric proportions determined in part (b) are related to the balanced equation for the reaction.

When the equation below is balanced, the correct set of stoichiometric coefficients is (a) \(1,6 \longrightarrow 1,3,4;\) (b) \(1,4 \longrightarrow 1,2,2 ;\) (c) \(2,6 \longrightarrow 2,3,2;\) (d) \(3,8 \longrightarrow 3,4,2\) \(\begin{aligned} ? \mathrm{Cu}(\mathrm{s})+? \mathrm{HNO}_{3}(\mathrm{aq}) & \longrightarrow ? \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+? \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+? \mathrm{NO}(\mathrm{g}) \end{aligned}\)

When water and methanol, \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{l}),\) are mixed, the total volume of the resulting solution is not equal to the sum of the pure liquid volumes. (Refer to Exercise 99 for an explanation.) When \(72.061 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) and \(192.25 \mathrm{g}\) \(\mathrm{CH}_{3} \mathrm{OH}\) are mixed at \(25^{\circ} \mathrm{C},\) the resulting solution has a density of \(0.86070 \mathrm{g} / \mathrm{mL} .\) At \(25^{\circ} \mathrm{C},\) the densities of water and methanol are \(0.99705 \mathrm{g} / \mathrm{mL}\) and \(0.78706\) \(\mathrm{g} / \mathrm{mL},\) respectively. (a) Calculate the volumes of the pure liquid samples and the solution, and show that the pure liquid volumes are not additive. [ Hint: Although the volumes are not additive, the masses are.] (b) Calculate the molarity of \(\mathrm{CH}_{3} \mathrm{OH}\) in this solution.

Silver nitrate is a very expensive chemical. For a particular experiment, you need \(100.0 \mathrm{mL}\) of \(0.0750 \mathrm{M}\) \(\mathrm{AgNO}_{3},\) but only \(60 \mathrm{mL}\) of \(0.0500 \mathrm{M} \mathrm{AgNO}_{3}\) is available. You decide to pipet exactly \(50.00 \mathrm{mL}\) of the solution into a \(100.0 \mathrm{mL}\) flask, add an appropriate mass of \(\mathrm{AgNO}_{3},\) and then dilute the resulting solution to exactly \(100.0 \mathrm{mL}\). What mass of \(\mathrm{AgNO}_{3}\) must you use?

Chalkboard chalk is made from calcium carbonate and calcium sulfate, with minor impurities such as \(\mathrm{SiO}_{2} .\) Only the \(\mathrm{CaCO}_{3}\) reacts with dilute \(\mathrm{HCl}(\mathrm{aq})\) What is the mass percent \(\mathrm{CaCO}_{3}\) in a piece of chalk if a 3.28 -g sample yields \(0.981 \mathrm{g} \mathrm{CO}_{2}(\mathrm{g}) ?\) $$\begin{aligned} \mathrm{CaCO}_{3}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) & \longrightarrow \mathrm{CaCl}_{2}(\mathrm{aq}) +\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g}) \end{aligned}$$
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