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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

Consider the chemical equation below. What is the maximum number of moles of \(\mathrm{K}_{2} \mathrm{SO}_{4}\) that can be obtained from a reaction mixture containing 5.0 moles each of \(\mathrm{KMnO}_{4}, \mathrm{KI},\) and \(\mathrm{H}_{2} \mathrm{SO}_{4} ?\) (a) \(3.0 \mathrm{mol}\); (b) \(3.8 \mathrm{mol}\) (c) \(5.0 \mathrm{mol} ;\) (d) \(6.0 \mathrm{mol} ;\) (e) \(15 \mathrm{mol}\). $$\begin{array}{r} 2 \mathrm{KMnO}_{4}+10 \mathrm{KI}+8 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow 6 \mathrm{K}_{2} \mathrm{SO}_{4}+2 \mathrm{MnSO}_{4}+5 \mathrm{I}_{2}+8 \mathrm{H}_{2} \mathrm{O} \end{array}$$

A seawater sample has a density of \(1.03 \mathrm{g} / \mathrm{mL}\) and \(2.8 \% \mathrm{NaCl}\)l by mass. A saturated solution of \(\mathrm{NaCl}\) in water is \(5.45 \mathrm{M} \mathrm{NaCl} .\) How many liters of water would have to be evaporated from \(1.00 \times 10^{6} \mathrm{L}\) of the seawater before \(\mathrm{NaCl}\) would begin to crystallize? (A saturated solution contains the maximum amount of dissolved solute possible.)

Consider the reaction below. \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{CaCl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) (a) How many grams of \(\mathrm{Ca}(\mathrm{OH})_{2}\) are required to react completely with \(415 \mathrm{mL}\) of \(0.477 \mathrm{M} \mathrm{HCl} ?\) (b) How many kilograms of \(\mathrm{Ca}(\mathrm{OH})_{2}\) are required to react with 324 L of a HCl solution that is 24.28\% HCl by mass, and has a density of \(1.12 \mathrm{g} / \mathrm{mL} ?\)

Aluminum metal and iron(III) oxide react to give aluminum oxide and iron metal. What is the maximum mass of iron that can be obtained from a reaction mixture containing \(2.5 \mathrm{g}\) of aluminum and \(9.5 \mathrm{g}\) of iron(III) oxide. What mass of the excess reactant remains?

A 0.3126 g sample of oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) requires 26.21 mL of a particular concentration of \(\mathrm{NaOH}(\mathrm{aq})\) to complete the following reaction. What is the molarity of the \(\mathrm{NaOH}(\mathrm{aq}) ?\) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{s})+2 \mathrm{NaOH}(\mathrm{aq}) \longrightarrow\) $$ \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) $$

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