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The feed to an ammonia synthesis reactor contains 25 mole \(\%\) nitrogen and the balance hydrogen. The flow rate of the stream is \(3000 \mathrm{kg} / \mathrm{h}\). Calculate the rate of flow of nitrogen into the reactor in \(\mathrm{kg} / \mathrm{h}\). (Suggestion: First calculate the average molecular weight of the mixture.)

Short Answer

Expert verified
The rate of flow of nitrogen into the reactor is found to be \( 3000 \, \mathrm{kg/h} \)

Step by step solution

01

Determine the Average Molecular weight

The average molecular weight of the mixture can be obtained by summing up the products of the mole fraction and the molecular weight of each component. The mole fraction of Nitrogen (N2) and Hydrogen (H2) are 25 and 75 \(\%\) respectively. Molecular weights for Nitrogen and Hydrogen are \(28 \, \mathrm{g/mol}\) and \(2 \, \mathrm{g/mol}\) respectively. \( M_{avg} = 0.25*28 + 0.75*2 = 7 \, \mathrm{g/mol} \)
02

Convert Total Feed Rate from Mass to Moles

The total feed rate in mass (3000 \, \mathrm{kg/h}\) is converted into moles using the average molecular weight. \(n_{total} = \frac{3000000 \, \mathrm{g/h}}{7 \, \mathrm{g/mol}} \approx 428571.43 \, \mathrm{mol/h}\)
03

Calculate the Nitrogen Flow Rate

Finally, the mole flow rate of Nitrogen can be determined by multiplying the total feed rate in moles by the mole fraction of Nitrogen. Later, this value is converted to mass per hour by multiplying with the molecular weight of nitrogen. \(n_{N2} = 0.25 * 428571.43 \, \mathrm{mol/h} = 107142.86 \, \mathrm{mol/h} \). Converting moles to mass gives, \( m_{\mathrm{N2}} = 107142.86 \, \mathrm{mol/h} * 28 \, \mathrm{g/mol} \approx 3000000 \, \mathrm{g/h} = 3000 \, \mathrm{kg/h}\)
04

Provide the Final Answer

The nitrogen flow rate into the reactor is 3000 kg/h

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Key Concepts

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

Mole Fraction Explained
Imagine you have a mixture of two or more substances. The mole fraction is a way of expressing the concentration of a particular component in that mixture. It's simply the ratio of the number of moles of the component of interest to the total number of moles of all components in the mixture.

The formula for mole fraction (\(X\textsubscript{i}\)) is:
\[\begin{equation}X_i=\frac{n_i}{n_{total}}\end{equation}\]
where \begin{itemize}\item \(n_i\) is the number of moles of the component, and\item \(n_{total}\) is the total number of moles of all components in the mixture.\end{itemize}
For instance, if you have a container filled with nitrogen and hydrogen gases, and 25 moles out of a total of 100 moles are nitrogen, the mole fraction of nitrogen would be 0.25 or 25%. It's a simple but fundamental concept in chemical engineering calculations for understanding the composition of mixtures.
Average Molecular Weight
To comprehend the average molecular weight, consider it as the 'weighted average' of the molecular weights of all components in a mixture. Each component's molecular weight is multiplied by its respective mole fraction. By adding these products together, we get the average molecular weight of the entire mixture.

The equation looks like this:
\[\begin{equation}M_{avg}=\sum(X_i \times M_i)\end{equation}\]
where
  • \(X_i\) is the mole fraction,
  • \(M_i\) is the molecular weight of each component,
  • the sum is over all the components in the mixture.

This concept is particularly useful when dealing with mixtures, as it allows for the conversion between mass and moles of a mixture rather than having to do so for each individual component.
Molar Mass Conversion
One of the most frequent conversions in chemical engineering is from mass to moles or vice versa. This is where molar mass conversion comes into play. The molar mass (or molecular weight) is the mass of one mole of a substance.

The conversion formula is:
\[\begin{equation}n=\frac{m}{M}\end{equation}\]
where
  • \(n\) is the number of moles,
  • \(m\) is the mass of the substance (in grams or kilograms),
  • \(M\) is the molar mass of the substance (in grams per mole or kilograms per mole).

This equation comes in handy when there's a need to convert the given mass of a feed stream, like in the ammonia synthesis example, into moles before proceeding with composition-based calculations.
Ammonia Synthesis Process
Ammonia synthesis refers to the industrial process used to produce ammonia (\(NH_3\)) from nitrogen (\(N_2\)) and hydrogen (\(H_2\)) gases. The heart of this production is the Haber-Bosch process, which combines nitrogen and hydrogen under high temperatures and pressures in the presence of a catalyst. The reaction can be simplified as:
\[\begin{equation}N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)\end{equation}\]
Understanding the mole fraction and average molecular weight, as discussed previously, is crucial in this process. They determine the right proportions of nitrogen to hydrogen and help in calculating the quantities involved in the reaction. Chemical engineers utilize these calculations to ensure the optimum synthesis of ammonia, which is a key ingredient for fertilizers and essential for agriculture.

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Most popular questions from this chapter

Certain solid substances, known as hydrated compounds, have well-defined molecular ratios of water to some other species. For example, calcium sulfate dihydrate (commonly known as gypsum, \(\left.\mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\right),\) has 2 moles of water per mole of calcium sulfate; alternatively, it may be said that 1 mole of gypsum consists of 1 mole of calcium sulfate and 2 moles of water. The water in such substances is called water of hydration. (More information about hydrated salts is given in Chapter 6 .) In order to eliminate the discharge of sulfuric acid into the environment, a process has been developed in which the acid is reacted with aragonite \(\left(\mathrm{CaCO}_{3}\right)\) to produce calcium sulfate. The calcium sulfate then comes out of solution in a crystallizer to form a slurry (a suspension of solid particles in a liquid) of solid gypsum particles suspended in an aqueous \(\mathrm{CaSO}_{4}\) solution. The slurry flows from the crystallizer to a filter in which the particles are collected as a filter cake. The filter cake, which is 95.0 wiff solid gypsum and the remainder CaSO_solution, is fed to a dryer in which all water (including the water of hydration in the crystals) is driven off to yield anhydrous (water-free) CaSO \(_{4}\) as product. A flowchart and relevant process data are given below. Solids content of slurry leaving crystallizer: \(0.35 \mathrm{kg} \mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O} / \mathrm{L}\) slurry \(\mathrm{CaSO}_{4}\) content of slurry liquid: \(0.209 \mathrm{g} \mathrm{CaSO}_{4} / 100 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) Specific gravities: \(\mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{s}), 2.32 ;\) liquid solutions, 1.05 (a) Briefly explain in your own words the functions of the three units (crystallizer, filter, and dryer). (b) Takea basis of one liter of solution leaving the crystallizer and calculate the mass (kg) and volume (L) of solid gypsum, the mass of \(\mathrm{CaSO}_{4}\) in the gypsum, and the mass of \(\mathrm{CaSO}_{4}\) in the liquid solution. (c) Calculate the percentage recovery of \(\mathrm{CaSO}_{4}-\) that is, the percentage of the total \(\mathrm{CaSO}_{4}\) (precipitated plus dissolved) leaving the crystallizer recovered as solid anhydrous \(\mathrm{CaSO}_{4}\) (d) List five potential negative consequences of discharging \(\mathrm{H}_{2} \mathrm{SO}_{4}\) into the river passing the plant.

In the manufacture of pharmaceuticals, most active pharmaceutical ingredients (APIs) are made in solution and then recovered by separation. Acetaminophen, a pain-killing drug commercially marketed as Tylenol", is synthesized in an aqueous solution and subsequently crystallized. The slurry of crystals is sent to a centrifuge from which two effluent streams emerge: ( 1 ) a wet cake containing 90.0 wt\% solid acetaminophen \((\mathrm{MW}=\) 151 g/mol) and 10.0 wt\% water (plus some acetaminophen and other dissolved substances, which we will neglect), and (2) a highly dilute aqueous solution of acetaminophen that is discharged from the process. The wet cake is fed to a dryer where the water is completely evaporated, leaving the residual acetaminophen solids bone dry. If the evaporated water were condensed, its volumetric flow rate would be \(50.0 \mathrm{Lh}\). Following is a flowchart of the process, which runs 24 h/day, 320 days/yr. A denotes acetaminophen. (a) Calculate the yearly production rate of solid acetaminophen (tonne/yr), using as few dimensional equations as possible. (b) A proposal has been made to subject the liquid solution leaving the centrifuge to further processing to recover more of the dissolved acetaminophen instead of disposing of the solution. On what would the decision depend?

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An object of density \(\rho_{\mathrm{a}},\) volume \(V_{\mathrm{a}},\) and weight \(W_{\mathrm{a}}\) is thrown from a rowboat floating on the surface of a small pond and sinks to the bottom. The weight of the rowboat without the jettisoned object is \(W_{\mathrm{b}}\). Beforethe object was thrown out, the depth of the pond was \(h_{\mathrm{pl}}\), and the bottom of the boat was a distance \(h_{\mathrm{b} 1}\) above the pond bottom. After the object sinks, the values of these quantities are \(h_{\mathrm{p} 2}\) and \(h_{\mathrm{b} 2}\). The area of the pond is \(A_{\mathrm{p}}\); that of the boat is \(A_{b} . A_{b}\) may be assumed constant, so that the volume of water displaced by the boat is \(A_{\mathrm{b}}\left(h_{\mathrm{p}}-h_{\mathrm{b}}\right)\). (a) Derive an expression for the change in the pond depth \(\left(h_{\mathrm{p} 2}-h_{\mathrm{p} 1}\right) .\) Does the liquid level of the pond rise or fall, or is it indeterminate? (b) Derive an expression for the change in the height of the bottom of the boat above the bottom of the pond \(\left(h_{b 2}-h_{b 1}\right) .\) Does the boat rise or fall relative to the pond bottom, or is it indeterminate?

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