RENDEMENT OF A REACTION
The
chemical rendement, the rendement of the reaction, or rendement refers to the
amount of reaction product produced in the chemical reaction. Absolute rendement can be written as weight in grams or in moles (molar yield). The relative yield used as a calculation of the effectiveness of the
procedure is calculated by dividing the amount of product obtained in
moles by the theoretical yield in moles: Fractional yield = actual rendement divided by theoretical rendement
To obtain a percentage yield, multiply the fractional yield by 100%.One or more reactants in chemical reactions are often used redundantly. The theoretical rendement is calculated based on the number of moles of the limiting reagent. For this calculation, it is usually assumed there is only one reaction involved. The ideal chemical yield value (theoretical rendement) is 100%, a value highly unlikely to be achieved in its practice. Calculate the percent of rendemen that is by using the following equations percent rendemen = weight yield / weight of yield divided by the sample weight multiplied by 100%
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In stoichiometry there is a special note onTheoretical results, tangible results, and percent of results
- Theoretical results are the amounts calculated from the balanced chemical equations or how many products are derived based on stoichiometric calculations.
- The real or actual result is the number of results actually generated in a chemical reaction or how many products are obtained after the reaction is complete.
- Efficiency of chemical reaction can be determined through percentage of result or percentage of rendemen that is: Percent of result: (result of real / theoretical result) x 100%
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The efficiency of a chemical reaction can be determined by calculating the percentage of results. Almost in all reactions, we will get fewer results than expected. This
happens because most of the reactions are equilibrium reactions (see:
Chemical Equilibrium) or because of some reaction conditions that cause
the reaction not to run perfectly. Chemists can obtain reaction efficiency by calculating the following percentage of results:Percentage of results = (actual results / theoretical results) x 100%
The real result is how many products are obtained after the reaction is complete. The theoretical result is how many products are obtained based on stoichiometric calculations. The comparison of these two results provides an explanation of how efficient the reaction is. In the previous example, the theoretical result of ferrous metals was 699.47 grams. While the real result is 525 grams. Therefore, the percentage of the results is:
% Yield = (525 grams / 699.47 grams) x 100% = 75.05%
The 75% yield percentage is not a too bad result. Chemical chemists and chemical engineers, however, prefer to earn more than 90%. One industry that uses the Haber Process has a yield percentage greater than 99%. In some chemical reactions, the reactants provided do not always correspond to their stoichiometric ratio. This means, we will run out of one of the reactants and still leave another reactant. The former reactant is known as a limiting reagent. The limiting reagent determines the amount of product to be produced by a chemical reaction. Here we will discuss how to determine the limiting reagents through the following example:
The real result is how many products are obtained after the reaction is complete. The theoretical result is how many products are obtained based on stoichiometric calculations. The comparison of these two results provides an explanation of how efficient the reaction is. In the previous example, the theoretical result of ferrous metals was 699.47 grams. While the real result is 525 grams. Therefore, the percentage of the results is:
% Yield = (525 grams / 699.47 grams) x 100% = 75.05%
The 75% yield percentage is not a too bad result. Chemical chemists and chemical engineers, however, prefer to earn more than 90%. One industry that uses the Haber Process has a yield percentage greater than 99%. In some chemical reactions, the reactants provided do not always correspond to their stoichiometric ratio. This means, we will run out of one of the reactants and still leave another reactant. The former reactant is known as a limiting reagent. The limiting reagent determines the amount of product to be produced by a chemical reaction. Here we will discuss how to determine the limiting reagents through the following example:
4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)
Mol NH3 = 100 gram / 17,024 gram.mol-1 = 5,874 mol
Mol NH3 / NH3 coefficient = 5,874 / 4 = 1,468
Mol O2 = 100 gram / 32.00 gram.mol-1 = 3.125 mol
Mol O2 / coefficient O2 = 3.125 / 5 = 0.625
The ammonia gas has a ratio of mole to the coefficient of 1.468. Meanwhile, oxygen gas has a value of ratio of 0.625. Thus, oxygen gas is a limiting reagent. The calculation of the product to be produced depends on the oxygen gas mole.
The stoichiometric ratio of
NO to O2 is 4: 5
Mol O2: Mol NO = reaction coefficient O2: reaction coefficient NO
3,125: Mole NO = 5: 4
Mol NO = 4/5 x Mol O2 = 4/5 x 3.125 mol = 2.5 mol NO
Mass NO = mole NO x Ar NO = 2.5 mole NO x 30.00 gram NO / mol NO = 75.00 gram NO
NO to O2 is 4: 5
Mol O2: Mol NO = reaction coefficient O2: reaction coefficient NO
3,125: Mole NO = 5: 4
Mol NO = 4/5 x Mol O2 = 4/5 x 3.125 mol = 2.5 mol NO
Mass NO = mole NO x Ar NO = 2.5 mole NO x 30.00 gram NO / mol NO = 75.00 gram NO
The value of 75.00 grams of NO is the theoretical result. If the actual result is 70.00 grams, the percentage of the reaction product is (70.00 gram / 75.00 gram) x 100% = 93.33%.We can also calculate how much ammonia gas is left. The mole gas calculations of ammonia used in the reaction depend on the mole of oxygen gas as a limiting reagent.
The NH3 stoichiometric ratio to O2 is 4: 5
Mol O2: Mol NH3 = reaction coefficient O2: The reaction coefficient NH3
3.125: Mol NH3 = 5: 4
Mol NH3 = 4/5 x Mol O2 = 4/5 x 3,125 mol = 2.5 mol of NH3
Mass NH3 = mol NH3 x Ar NH3 = 2.5 mol NH3 x 17.024 grams NH3 / mol NH3 = 42.56 grams NH3Thus, the amount of ammonia gas remaining (not used) is as much as 100 grams - 42.56 grams = 57.44 grams.
Mol O2: Mol NH3 = reaction coefficient O2: The reaction coefficient NH3
3.125: Mol NH3 = 5: 4
Mol NH3 = 4/5 x Mol O2 = 4/5 x 3,125 mol = 2.5 mol of NH3
Mass NH3 = mol NH3 x Ar NH3 = 2.5 mol NH3 x 17.024 grams NH3 / mol NH3 = 42.56 grams NH3Thus, the amount of ammonia gas remaining (not used) is as much as 100 grams - 42.56 grams = 57.44 grams.
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Problems example:
If 204 grams of bauxite (Mr. Al2O3 = 102) is perfectly reduced yields 54 grams of aluminum metal (Ar Al = 27). The percentage of Al2O3 in the trailer is ...
Answer:
In this case we are asked to calculate the percentage of Al2O3 in the trailer. This means the trailer is not pure containing 100% bauxite. So, our task in solving this problem is to seek the purity of the bauxite footage. To do this we must also compare FACT with THEORY
FACT : Al periods produced from 204 grams of Al2O3 compounds
THEORY : mass Al in Al2O3 = (2x27) : 102 = 108 grams
% Al2O3 = (fact : theory) x 100% = (54 : 108) x 100% = 50%
so, result or percentage of rendemen of Al2O3 is 50%
What is the difference between the actual results with theoretical results?
BalasHapusThe real result is how many products are obtained after the reaction is over. Whereas the theoretical result is how many products are obtained based on stoichiometric calculations. For example, in an experiment, the theoretical result of ferrous metals was 699.47 grams, while the real result is 525 grams.
HapusMentioned the reason why in a chemical reactions, the results of who had so much smaller than I expected
BalasHapusThere are several reasons why the results obtained from a chemical experiment are much smaller than expected. The reason is that the reactants used are not pure, the reaction technique used is not so good. Or maybe, it does not rule out that the reaction is an equilibrium reaction so we will never get 100% the same result as the theory
HapusMention and explain 1 example of the application of the concept of rendemen
BalasHapusExamples of the application of the concept of rendement in everyday life is the rendement of oil asiri
HapusRendement is the ratio of the amount (quantity) of oil produced from the extraction of aromatic plants. Rendemen use percent unit (%). The higher the yield value indicates the value of the oil produced more and more. Increasing the yield or comparison of the amount of oil produced can be done with two approaches:
First, the process of cultivation.
Second, the process of making oil
The quality of oil produced is usually inversely proportional to the amount of rendement produced. The higher the value of the resulting rendement the lower the quality obtained. The formula to calculate the rendement as follows:
Percentage rendement = (amount of oil produced: amount of material before processed) x 100%
Each plant has a different rendement value, for example ginger 1.44 to 2% while turmeric 7.8%
If the reaction between 6 grams of ethane C2H6 (Mr = 30) with 7.1 grams of Cl2 (Mr = 71) yields 5.16 grams of C2H6Cl (Mr = 64.5) then the percentage of ethylchloride rendement is ...
BalasHapusI think the result of the reaction is not C2H6Cl but C2H5Cl
Hapusfact:
The resulting mass of C2H5Cl is 5.16 grams. Then the mole is 0.08.
theory:
We look for the mass of C2H5Cl produced from the reaction
Mol C2H6 is 0.2 and mole Cl2 is 0.1
The reaction is happening
2C2H6 + Cl2 ---> 2C2H5Cl + H2
Firs : 0.2 mol 0.1 mol
Reaction: 0.2 mol 0.1 mol 0.2 mol 0.1 mol
Residual: - - 0.2 mol 0.1 mol
Theoretically C2H5Cl resulting from the reaction is 0.2 mol. So the percentage of yield of C2H5Cl formed is (0,08: 0,2) x 100% = 40%
Please explain again what is meant by rendemen and give an analogy of everyday life about the meaning of the rendemen
BalasHapusRendement is actually a term in the field of chemistry studies. Rendemen describes the inaccuracy of the reaction product, where the reaction result is always lower than the mathematical calculation (theory). For example, in a chemical reaction, it should produce a 100 gram (in theory) substance, but in reality the result is only 90 grams.
HapusUnconsciously this also often happens in everyday life. For example you target on the test will get the value of 80, but in fact you get the value of 70 or 60. Another analogy is in the field of sales, for example a sales are given the target must sell 10 products, but the product sold only 8. that's the analogy of understanding rendement
Please explain aboutDouble Displacement Reactions
BalasHapusEach reaction has reactants that react with each other to form a compound called a product. In most cases, the reactants are compounds. Each of these reacting compounds lends a single molecule or compound to exchange with molecules or compounds offered by other compounds.
HapusAnions and cation compounds change their placement to react and form completely different compounds. This type of reaction is also called metasynthesis or multiple replacement.
The standard representation of the type of equation is:
UV + XY ➜ UY + XV
This equation can be represented in terms of ion costs as follows:
U + V + X + Y - U + Y - + X + V -
Let us understand the features of this type of reaction with chemical equations in which silver nitrate reacts with sodium chloride to form the resultant products-sodium nitrate and silver chloride.
AgNO3 + NaCl ➜ NaNO3 + AgCl
In standard representations, U, V, X, and Y are compounds or elements. In the given example, AgNO3 consists of silver and nitrate. Nitrates, in turn, are nitrogen and oxygen compounds. NaCl contains only two elements, sodium and chlorine.
Reactants or compounds react with each other ionic in nature. In the given example, Cl- and NO3- are negative ions.
There is an exchange of elements with each other. Silver nitrate provides far NO3- and sodium chloride gives away Cl-. These ions are exchanged by two compounds, forming sodium nitrate and silver chloride.
The fact that silver nitrate is a compound containing three elements. All of the NO3 ions will be attached to the sodium atom.
The reactant of this equation is in aqueous form. However, the product may be precipitate, gas, or liquid. If the product is obtained in a gas or solid form, it becomes easier to separate.
Examples of double-boosting reactions are neutralization reactions, precipitation reactions and gaseous reactions