what is experimental yield

The actual yield is a product that is obtained by experimentation. The theoretical yield is obtained through stoichiometric calculation.

If the two yields are equal, you have 100 % yield.

Usually you obtain less than 100 %.

It is impossible to obtain a yield greater than 100%. If that happens, some experimental error must have occurred in the creation of the desired product, actual yield.

Actual Yield Calculator

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This actual yield calculator will answer your questions about how to calculate the actual yield and help you find the true yield of a chemical reaction. Because reality is much different than theory, no chemical reaction is ever carried out flawlessly. Because of this, the actual yield definition states that it should always be lower than the theoretical yield. Before commencing any laboratory work, figure out what the theoretical yield is so you know how much of your product to expect from a given amount of starting material. The efficiency of your reaction will help you compute the actual yield of a chemical process using the actual yield formula - or you can simply use our amazing tool!

Remember, that we also have a theoretical yield calculator and a percent yield calculator available for your chemistry calculations!

What is actual yield? Actual yield definition

The quantity of a product received from a chemical reaction is known as the actual yield. The theoretical yield, on the other hand, is the quantity of product that could be obtained if all of the reactants were converted to product perfectly, i.e., there were no by-products. The limiting reactant is used to calculate theoretical yield. By determining the percent yield, also known as the efficiency of the chemical reaction, you can calculate the actual yield using our actual yield calculator.

How to calculate the actual yield? Actual yield formula

Now that we know the actual yield definition, let's proceed to the actual yield formula for a chemical process:

• Y a Y_{\text{a}} Y a ​ — Actual yield;
• Y p Y_{\text{p}} Y p ​ — Percent yield ( % \% % ); and
• Y t Y_{\text{t}} Y t ​ — Theoretical yield.

Difference between theoretical yield and actual yield

The actual yield is usually lower than the theoretical yield because few reactions proceed to absolute completion (i.e., they aren't 100% efficient) or because not all of the product in a reaction is recovered. For example,

• If you're recovering a precipitate, you can lose some of it if it doesn't completely crash out of the solution;
• Some products may remain on the filter paper or make their way through the mesh and wash away if you filter the solution through filter paper;
• Even if the substance is insoluble in a particular solvent, if you rinse it, a little bit of it may be lost due to dissolving in the solvent; and
• Sometimes your product may undergo unwanted reactions after you have formed it due to the presence of high temperatures or other chemicals.

But, it is also possible that the actual yield exceeds the theoretical yield:

• This happens most frequently when the solvent is still present in the product (incomplete drying), when the product is weighed incorrectly, or when an unaccounted ingredient in the reaction acts as a catalyst or causes the creation of by-products; and
• Another reason for the increased yield is that the result is impure because another ingredient than the solvent is present.

Examples of how to find actual yield

Let's move on to real-world applications now that we've learned about the distinctions between actual and theoretical yields.

Example 1 : If the percent yield of a reaction was 45 % 45\% 45% with a theoretical yield of Y t = 4  g Y_{\text{t}} = 4\ \text{g} Y t ​ = 4   g , what is the actual yield?

You can solve this problem by using our actual yield calculator:

Input the percent yield: Y p = 45 % Y_{\text{p}} = 45\% Y p ​ = 45% .

Input the theoretical yield: Y t = 4  g Y_{\text{t}} = 4\ \text{g} Y t ​ = 4   g .

The actual yield is Y a = 1.8  g Y_{\text{a}}=1.8\ \text{g} Y a ​ = 1.8   g .

Example 2 : When making a new substance from other substances, chemists say that they have synthesized a new compound. Reactants are converted to products, and the process is symbolized by a chemical equation. For example, the decomposition of magnesium carbonate to form magnesium oxide has an experimental percent yield of 79 % 79\% 79% . The theoretical yield is known to be Y t = 19  g Y_{\text{t}}=19\ \text{g} Y t ​ = 19   g . What is the actual yield of magnesium oxide?

So, how to find the actual yield? The calculation is simple if you know the percent and theoretical yields. All you need to do is plug the values into the calculator again:

Input the percent yield: Y p = 79 % Y_{\text{p}} = 79\% Y p ​ = 79% .

Enter the theoretical yield: Y t = 19  g Y_{\text{t}} = 19\ \text{g} Y t ​ = 19   g .

The actual yield is Y a = 15  g Y_{\text{a}}=15\ \text{g} Y a ​ = 15   g .

Usually, you have to calculate the theoretical yield based on the balanced equation. In this equation, the reactant and the product have a 1 : 1 1:1 1 : 1 mole ratio, so, if you know the amount of reactant, you know the theoretical yield is the same value in moles (not grams!). You take the number of grams of reactant you have, convert it to moles (maybe with our grams to moles calculator ), and then use this number of moles to find out how many grams of product to expect.

You can find the yield of the reaction at every moment using the concentrations you can calculate with our reaction quotient calculator !

🙋 Make sure the theoretical yield and actual yield are expressed in the same unit: if you need to know the conversion, visit our weight converter .

How do I find actual yield?

To find the actual yield , simply follow these steps:

Use the actual yield formula:

Ya = (Yp /100) × Yt

Here Ya is the actual yield, Yp is the percent yield, and Yt is the theoretical yield.

Substitute the values for percent and theoretical yield.

That's it! With these two values, you can easily calculate the actual yield of a chemical reaction.

How do I find actual yield without percent yield?

To find the actual yield without percent yield , perform an experiment and weigh the product. To verify the accuracy of your measurement, you can calculate the efficiency or percent yield using the theoretical yield, which you can obtain from the reaction's stoichiometry. Percent yields below 70% usually indicate errors in the experiment or calculations.

Is actual yield the same as percent yield?

No , they are different. Actual yield refers to the product's weight produced in an experiment. In contrast, percent yield or reaction efficiency indicates how much the actual yield deviates from the theoretical amount of product expected.

What's the actual yield of CO₂ in the combustion of 14g of CH₄?

Assuming a percent yield Yp of 70% , the actual yield of CO 2 is 26.91 g :

Calculate the moles of CH 4 from its molar mass:

moles_CH 4 = 14g / 16.04 g/mol = 0.873 mol

Balance the combustion reaction:

CH 4 + 2O 2 -> CO 2 + 2H 2 O

Note that 1 mole of CH 4 produces 1 mole of CO 2 .

Calculate the theoretical yield Yt of CO 2 :

Yt_CO 2 = moles_CH 4 × molar_mass_CO 2

Yt_CO 2 = 0.873 mol × 44.01 g/mol = 38.44 g

Determine the actual yield Ya of CO 2 :

Ya_CO 2 = Yp × Yt_CO 2 Ya_CO 2 = 0.7 × 38.44 g = 26.91 g

Theoretical yield

Percent yield

Actual yield

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Percent Yield Formula and Definition

In chemistry, percent yield is a comparison of actual yield to theoretical yield , expressed as a percentage . Here is a look at the percent yield formula, how to calculate it, and why it may be less than or greater than 100%.

Percent Yield Formula

The percent yield formula is actual yield divided by theoretical yield in moles, multiplied by 100%:

Percent Yield = Actual Yield/Theoretical Yield x 100%

It does not matter whether you express actual and theoretical yield in grams or moles , as long as you use the same units for both values.

How to Calculate Percent Yield

Calculating percent yield requires two values: the actual yield and the theoretical yield. Yield depends on the mole ratio between reactants and products . The actual yield is the amount of product obtained from a reaction or experiment. You weigh the product and then convert the mass (usually in grams) to moles .

The theoretical yield comes from stoichiometry. In other words, it comes from the mole ratio between reactants and products in the balanced equation for the chemical reaction. Once you have the balanced equation, the next step is finding the limiting reactant. The limited reactant is the reactant that limits the amount of product because it’s consumed before the other reactant runs out. In a decomposition reaction, there may only be one reactant, which makes it the limiting reactant. In other reactions, you compare the molar masses and mole ratios. Next, use the number of moles of limiting reactant and mole ratio and calculate the theoretical yield. Finally, calculate theoretical yield.

• Balance the chemical equation for the reaction. Note the number of moles of reactants and products.
• Identify the limiting reactant. Calculate molar masses of all reactants. Find the number of moles of all species. Use the mole ratio and identify which reactant limits the reaction. Use the number of moles of the limiting reactant and mole ratio and find the theoretical yield.
• Weigh the product. This is actual yield.
• Make sure actual yield and theoretical yield have the same units (grams or moles).
• Calculate percent yield using actual yield and theoretical yield.

Example Percent Yield Calculation (Simple)

First, here is a simple example of the percent yield calculation in action:

The decomposition of magnesium carbonate forms 15 grams of magnesium oxide in an experiment. The theoretical yield is 19 grams. What is the percent yield of magnesium oxide?

MgCO 3  → MgO + CO 2

Here, you know the actual yield (15 grams) and the theoretical yield (19 grams), so just plug the values into the formula:

Percent Yield = Actual Yield/Theoretical Yield x 100% Percent Yield = 15 g/19 g x 100% Percent Yield = 79%

Example Percent Yield Calculation (With Limiting Reactant)

Find the percent yield of a reaction when you obtain 4.88 g of AlCl 3 (s) from a reaction between 2.80 g Al(s) and 4.15 g Cl 2 (g).

First, write out the balanced equation for the reaction:

2Al( s ) + 3Cl 2 ​( g ) → 2AlCl 3 ​( s )

Next, find the limiting reactant. Start with the molar masses of the reactants and products:

2.80 g Al x (1 mol Al/26.98 g Al) = 0.104 mol Al 4.15 g Cl 2 x (1 mol Cl 2 /70.90 g Cl 2 ) = 0.0585 mol Cl 2

Compare the mole ratio to the actual number of moles present in the reaction. From the balanced equation, you see 2 moles of Al reacts with 3 moles of Cl 2 .

Mole ratio: moles Al/moles Cl 2 = 2/3 = 0.6667 Actual ratio moles Al/moles Cl 2 = 0.104/0.0585 = 1.78

The actual ratio is larger than the mole ratio, so there is excess Al and Cl 2 is the limiting reactant. (If the actual ratio is smaller than the mole ratio, it means there is excess Cl 2 and Al is the limiting reactant.)

Use the actual number of moles of Cl 2 and the mole ratio and find the maximum amount of AlCl 3 .

0.00585 mol Cl 2 x (2 mol AlCl 3 /3 mol Cl 2 ) = 0.00390 mol AlCl 3

Convert the number of moles of product into grams so the units of actual and theoretical yield are the same. Get this from the molar mass.

0.00390 mol AlCl 3 x (133.33 g AlCl 3 /1 mol AlCl 3 ) = 5.20 g AlCl 3

Finally, calculate percent yield. Actual yield is 4.88 g of AlCl 3 (given in the problem) and theoretical yield is 5.20 g AlCl 3 .

Percent Yield = Actual Yield/Theoretical Yield x 100% Percent Yield = 4.88 g AlCl 3 / 5.20 g AlCl 3 x 100% Percent Yield = 93.8%

Is Percent Yield Always Less Than 100%

Percent yield is always less than 100% (often by a lot), yet it’s possible to calculate a value greater than 100%.

There are a few reasons why percent yield always falls short.

• Not all reactions proceed to completion.
• Sometimes reactants and products exist in equilibrium, so the reverse reaction also occurs.
• Two or more reactions occur simultaneously, converting some reactant to one or more side products.
• There may be other species or impurities that interfere with the reaction.
• Product gets lost during transfer.
• Product gets lost during purification.

And yet, sometimes you get more product than predicted. Occasionally, an impurity contributes to product formation. But, usually there’s less product than the theoretical yield. Yet, if you collect an impure product, the mass exceeds the theoretical yield. The most common situation is weighing product that isn’t completely dry. Part of the mass is solvent , so it appears you got more product than predicted.

• Cornforth, J. W. (1993). “The Trouble With Synthesis”. Australian Journal of Chemistry . 46 (2): 157–170. doi: 10.1071/ch9930157
• Petrucci, Ralph H.; Herring, F. Geoffrey; Madura, Jeffry; Bissonnette, Carey; Pearson (2017). General Chemistry: Principles and Modern Applications . Toronto: Pearson. ISBN 978-0-13-293128-1.
• Whitten, Kenneth W.; Davis, Raymond E; Peck, M. Larry (2002). General Chemistry . Fort Worth: Thomson Learning. ISBN 978-0-03-021017-4.
• Vogel, Arthur Israel; Furniss, B. S; Tatchell, Austin Robert (1978). Vogel’s Textbook of Practical Organic Chemistry . New York: Longman. ISBN 978-0-582-44250-4.

Related Posts

Percent Yield

Percent yield refers to the percent ratio of actual yield to the theoretical yield. In chemistry, yield is a measure of the quantity of moles of a product formed in relation to the reactant consumed, obtained in a chemical reaction, usually expressed as a percentage. The amount of product actually made compared with the maximum calculated yield is called the percentage yield. Let us understand the percent yield formula using solved examples.

What is Percent Yield?

In chemistry, percent yield is the percent ratio of the weight of the product obtained to the theoretical yield. We calculate the percent yield by dividing the experimental yield by the theoretical yield and multiplying the result by 100 to express the final answer in %. Generally, the value of percent yield is lower than 100%, since the actual yield obtained after the reaction is often less than the theoretical value. This can be because of an incomplete reaction.

Percent yield can also be greater than 100%, meaning that more sample was recovered from the reaction than the initially predicted yield. Percent yield is always positive.

Percentage Yield Example

Let us see the decomposition reaction of magnesium oxide to understand the concept of percent yield better.

MgCO\(_3\) → MgO + CO\(_2\)

This means for 1 mole of reactant[MgCO\(_3\)], we will obtain 1 mole of product[MgO], i.e, the reactant and the product have a 1:1 mole ratio. When given the amount of reactant, we can find the theoretical yield by comparing the mole ratio. The ratio of actual yield given to this theoretical yield gives the percentage yield.

Percent Yield Formula

The percentage yield formula is calculated to be the experimental yield divided by theoretical yield multiplied by 100. If the actual and theoretical yield ​is the same, the percent yield is 100%. Usually, the percent yield is lower than 100% because the actual yield is often less than the theoretical value.

Percent Yield = (Actual Yield / Theoretical Yield) × 100 %

Formula to Calculate Percent Yield

The formula to calculate the percent yield is:

Percentage Yield = (Actual Yield / Theoretical Yield) × 100 %

• Actual yield - it gives the amount of product obtained from a chemical reaction
• Theoretical yield - it gives the amount of product obtained from the stoichiometric or balanced equation, using the limiting reactant to determine the product
• Units for both actual and theoretical yield need to be the same (moles or grams)

Examples on Percent Yield

Example 1: During a chemical reaction, 0.5 g of product is made. The maximum calculated yield is 1.6 g. What is the percent yield of this reaction?

We know that according to Percent Yield Formula,

Percentage yield = (Actual yield/Theoretical yield)× 100%

= 0.5/1.6× 100%

Therefore, the percentage yield of this reaction is 31.25%.

Example 2: During a chemical reaction 1.8 g of product is made. The maximum calculated yield is 3.6 g. What is the percent yield of this reaction?

= 1.8/3.6× 100%

Therefore, the percentage yield of this reaction is 50%.

Example 3: If the percentage yield is 45% with the theoretical yield as 4g, what would the actual yield be? Calculate using the percentage yield formula.

Using the percentage yield formula,

45 = Actual yield/4 × 100

Actual yield = 1.8

Therefore, the actual yield is 1.8g

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Practice Questions on Percent Yield

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FAQs on Percent Yield

What is meant by percent yield.

Percent yield is a measure of the actual number of moles obtained for any reactant in any reaction in comparison to the predicted or theoretical yield.

What is Meant by Percent Yield Formula?

The percent yield formula is the percent ratio of actual yield to the theoretical yield. Yield is a measure of the quantity of moles of a product formed in relation to the reactant consumed, obtained in a chemical reaction, usually expressed as a percentage. The amount of product actually made compared with the maximum calculated yield is called the Percent yield. The formula is (Actual Yield / Theoretical Yield) × 100 %

What is the Formula to Calculate the Percent Yield?

The formula to calculate the percentage yield is:

Why do Use the Percent Yield Formula?

The percent yield is also used in the field of chemistry, where the percentage yield of a chemical reaction is considered very important. The formula to calculate the percentage yield is to compare the yield or the quantity of the product that is obtained.

Why is the Percent Yield not 100% When the Formula is Used?

In most cases, the percent yield is less than 100% because the actual yield is often less than the theoretical value. This is due to the incomplete or competing reactions and loss of samples during recovery.

Actual Yield Definition (Chemistry)

Actual Yield Versus Theoretical Yield

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Actual Yield Definition

The actual yield is the quantity of a product that is obtained from a chemical reaction . In contrast, the calculated or theoretical yield  is the amount of product that could be obtained from a reaction if all of the reactant converts to product. Theoretical yield is based on the limiting reactant .

Common Misspelling: actual yeild

Why Is Actual Yield Different from Theoretical Yield?

Usually, the actual yield is lower than the theoretical yield because few reactions truly proceed to completion (i.e., aren't 100% efficient) or because not all of the product in a reaction is recovered. For example, if you are recovering a product that is a precipitate, you may lose some product if it doesn't completely fall out of solution. If you filter the solution through filter paper, some product may remain on the filter or make its way through the mesh and wash away. If you rinse the product, a small amount of it may be lost from dissolving in the solvent, even if the product is insoluble in that solvent.

It's also possible for the actual yield to be more than the theoretical yield. This tends to occur most often if solvent is still present in the product (incomplete drying), from error weighing the product, or perhaps because an unaccounted substance in the reaction acted as a catalyst or also led to product formation. Another reason for higher yield is that the product is impure, due to the presence of another substance besides the solvent.

Actual Yield and Percent Yield

The relationship between actual yield and theoretical yield is used to calculate percent yield :

percent yields = actual yield / theoretical yield x 100%

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Experimental device for the “green” synthesis of unbranched aliphatic esters c4–c8 using an audio frequency electric field.

1. Introduction

2. results and discussions, 2.1. the experimental device based on green technology, 2.2. synthesis of c4–c8 acetate esters using an audio frequency electric field, 2.3. c4–c8 ester analysis using gas chromatography techniques, 2.4. mechanism of esterification using an audio frequency electric field, 2.5. agree analysis, 2.6. strengths, limitations, and future perspectives, 3. materials and methods, 3.1. the experimental device, 3.2. synthesis of aliphatic esters using an audio frequency electric field, 3.3. gc-fid analysis, 3.4. gc-ms analysis, 3.5. evaluation of preparation and analytical methods using agree tools, 4. conclusions, supplementary materials, author contributions, institutional review board statement, informed consent statement, data availability statement, acknowledgments, conflicts of interest.

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Udrea, I.-A.; Lukinich-Gruia, A.T.; Paul, C.; Pricop, M.-A.; Dan, M.; Păunescu, V.; Băloi, A.; Tatu, C.A.; Vaszilcsin, N.; Ordodi, V.L. Experimental Device for the “Green” Synthesis of Unbranched Aliphatic Esters C4–C8 Using an Audio Frequency Electric Field. Processes 2024 , 12 , 1891. https://doi.org/10.3390/pr12091891

Udrea I-A, Lukinich-Gruia AT, Paul C, Pricop M-A, Dan M, Păunescu V, Băloi A, Tatu CA, Vaszilcsin N, Ordodi VL. Experimental Device for the “Green” Synthesis of Unbranched Aliphatic Esters C4–C8 Using an Audio Frequency Electric Field. Processes . 2024; 12(9):1891. https://doi.org/10.3390/pr12091891

Udrea, Ioan-Alexandru, Alexandra Teodora Lukinich-Gruia, Cristina Paul, Maria-Alexandra Pricop, Mircea Dan, Virgil Păunescu, Alexandru Băloi, Călin A. Tatu, Nicolae Vaszilcsin, and Valentin L. Ordodi. 2024. "Experimental Device for the “Green” Synthesis of Unbranched Aliphatic Esters C4–C8 Using an Audio Frequency Electric Field" Processes 12, no. 9: 1891. https://doi.org/10.3390/pr12091891

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Ammonia Synthesis via Membrane Dielectric-Barrier Discharge Reactor Integrated with Metal Catalyst

• Original Paper
• Published: 03 September 2024

Cite this article

• Visal Veng 1 ,
• Saleh Ahmat Ibrahim 2 ,
• Benard Tabu 1 ,
• Ephraim Simasiku 1 ,
• Joshua Landis 1 ,
• John Hunter Mack 1 ,
• Fanglin Che 2 &
• Juan Pablo Trelles 1  

The synthesis of ammonia using non-thermal plasma can present distinct advantages for distributed stand-alone operations powered by electricity from renewable energy sources. We present the synthesis of ammonia from nitrogen and hydrogen using a membrane Dielectric-Barrier Discharge (mDBD) reactor integrated with metal catalyst. The reactor used a porous alumina membrane as a dielectric-barrier and as a distributor of H 2 , a configuration that leads to greater NH 3 production than using pre-mixed N 2 and H 2 . The membrane is surrounded by catalyst powder held by glass wool as porous dielectric support filling the plasma region. We evaluated nickel, cobalt, and bimetallic nickel-cobalt as catalysts due to their predicted lower activation energy under non-thermal plasma conditions as determined through Density Functional Theory (DFT) calculations. The catalysts were loaded at 5% by weight on alumina powder. The performance of the catalytic mDBD reactor was assessed using electrical, optical, and spectroscopic diagnostics, as well as Fourier-Transform Infrared spectroscopy. Experimental results showed that the glass wool support suppresses microdischarges, generally leading to greater ammonia production. The Ni-Co/Al 2 O 3 catalyst produced the greatest energy yield of 0.87 g-NH 3 /kWh, compared to a maximum of 0.82 and 0.78 g-NH 3 /kWh for the Co/Al 2 O 3 and Ni/Al 2 O 3 catalysts, respectively. Although the differences in performance among the three metal catalysts are small, they corroborate the predictions by DFT. Moreover, the maximum energy yield for bare Al 2 O 3 (no metal catalyst) with dielectric support was 0.38 g-NH 3 /kWh, for mDBD operation with no metal catalyst or dielectric support was 0.28 g-NH 3 /kWh, and for standard DBD operation (no membrane, dielectric support, or catalyst) was 0.08 g-NH 3 /kWh, i.e., 2.1, 3.1, and 11 times lower, respectively, than the maximum energy yield for the Ni-Co/Al 2 O 3 catalyst with dielectric support. The study shows that the integration of dielectric membrane and metal catalyst is an effective approach at enhancing ammonia production in a DBD reactor.

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Acknowledgements

This work was support by Department of Navy award N00014-22-1-2001 issued by the Office of Naval Research. The United States Government has a royalty-free license throughout the world for all copyrightable material contained herein. This research used the resources of the National Energy Research Scientific Computing Center (NERSC), a U.S. Department of Energy Office of Science User Facility supported by the Office of Science of the U.S. Department of Energy under Contract No. DE-AC02-05CH11231 using NERSC awards BES-ERCAP0027465 and BES-ERCAP0028368.

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Department of Mechanical and Industrial Engineering, University of Massachusetts Lowell, Lowell, MA, USA

Visal Veng, Benard Tabu, Ephraim Simasiku, Joshua Landis, John Hunter Mack & Juan Pablo Trelles

Department of Chemical Engineering, University of Massachusetts Lowell, Lowell, MA, USA

Saleh Ahmat Ibrahim & Fanglin Che

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Contributions

Visal Veng: Investigation, Visualization, Writing – Original Draft Preparation (lead); Saleh Ahmat Ibrahim: Density Functional Theory (DFT) calculations (supporting); Benard Tabu: optical emission spectroscopy (supporting); Ephraim Simasiku: Membrane DBD reactor design analysis (supporting); Joshua E. Landis: Resources: Fourier-Transform Infrared Spectroscopy (FTIR) analysis (supporting); J. Hunter Mack: Resources and Supervision: Fourier-Transform Infrared Spectroscopy (FTIR) analysis (lead); Fanglin Che: Density Functional Theory (DFT) calculations (lead); Juan Pablo Trelles: Conceptualization (lead), Funding Acquisition (lead), Project Administration (lead), Supervision (lead), and Writing – Review & Editing (lead).

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Veng, V., Ibrahim, S.A., Tabu, B. et al. Ammonia Synthesis via Membrane Dielectric-Barrier Discharge Reactor Integrated with Metal Catalyst. Plasma Chem Plasma Process (2024). https://doi.org/10.1007/s11090-024-10502-7

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Received : 05 February 2024

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Published : 03 September 2024

DOI : https://doi.org/10.1007/s11090-024-10502-7

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