LAB 4 CHM420 Stoichiometry and Theoretical Yield PDF

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lab report chm420
experiment 4 Stoichiometry and Theoretical Yield
objective, introduction, result, discussion, conclusion, reference...

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EXPERIMENT 4: Stoichiometry and Theoretical Yield

OBJECTIVE: 1. To identify the limiting reactant and excess reactant 2. To determine the percent yield of the product

INTRODUCTION

Chemical equations represent how a chemical reaction proceeds from reactants to products through physical or chemical change using chemical formulas. Stoichiometry method is used in order to solve the chemical equations of related problems. One can use this method to find the number of moles, mass and percent in chemical equations. Balancing an equation is an iterative process that requires adding coefficients to each side until the numbers become equal. There are several approaches to balance a chemical equation. For example, the method of stoichiometry is used in order to identify the limiting reactant, excess reactant and to find the percentage yield. Balancing the equation is also essential for determining the limiting reactant because the coefficient of the compounds is used to calculate how much product is produced by each reactant (product yield). From this quantity, the reactant producing the least amount of product is considered the limiting reactant, which is completely consumed in the reaction and therefore limits the total amount of product generated. This calculated quantity also represents the theoretical yield of the reaction, which is needed to calculate the percent yield.

The balanced equation is more than a simple accounting of atoms. The coefficients describe the molar relationship between products and reactants, example: how much product is produced by each reactant. The number of moles of reactant is used to calculate the number of moles of another product or reactant. The reactant that produces the least amount of product is considered the limiting reactant.The limiting reactant is completely consumed in the reaction and therefore limits the total amount of product generated. Once the limiting reactant is entirely consumed, no more product will form. The possible amount of product that could be formed based on the limiting reactant is the theoretical yield of the reaction.The actual yield is compared to the theoretical yield, resulting in the ‘percent yield’. A percent yield of 100% means that, based on the reactants used, the maximum possible amount of product was produced. Percent yields less than 100% are common and indicate that there was some product loss during the reaction. The percent yield is never greater than the theoretical yield. If this is the case, experimental or calculation errors occurred.

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RESULTS

Reaction 1

Reaction 2

Mass of the product, g (after 1st heating)

0.888

0.562

Mass of the product, g (after 2nd heating)

0.887

0.500

Mass of the product, g (after 3rd heating)

0.887

0.558

Mass of the product, g

0.8873

0.560

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DISCUSSION In this experiment of reacting 2 different volumes of reactant which prove that the yield of the product depends on the least volume of reactant. Thereby, two experiments are performed that used the same molarity of reactant but differ in volume. In reaction one the volume for 0.50M CaCl2 is 20.00ml and react with 10.00mL of 1.50M NaCo3 as for the reaction two the experiment used the same amount and molarity of CaCl2 in reaction one but 1.50M NaCO3 used 5.00mL. Therefore, both reactions produce a mass of Caco3 precipitate, the mass of precipitate is depending on the limiting reactant, this cause by the principle stoichiometry gives a quantitative relation between the number of atoms in both sides. Therefore , the stoichiometry balanced equation

can help to predict the mass of CaCO3 products by

mathematically calculating the ratio of number of mole from products and the limiting reactant. The limiting reactant is important because it controls the amount of product made when the reactant is totally consumed in the completed reaction and the reaction cannot react without it.

The experiment of mixing NaCO3 and CaCl2 solution produces the CaCO3 precipitate along with aqueous NaCl. The precipitate had accumulated and dried to determine the mass by weight. As a result, the mass of the product that is dependent by the limiting reactant had been recorded three times to increase the reliability of data. Based on the result in reaction one, the average of mass of CaCl3 is 0.8873g and for reaction two is 0.560g. The reliability of this experiment value has been determined by the percent yield between the ratio of this experimentally value and theoretical value. The theoretical value can be calculated by balancing the chemical equation to obtain the ratio of the number of moles and using formula to get the theoretical amount of product that should be produced from the limiting reactant. In this case, the limiting reactant for reaction 1 is CaCl2 and for reaction 2 is NaCO3, this is due to the number of moles between the both reactants which is smaller. Therefore the percentage value for the reaction 1 is 88.65% and for reaction 2 is 74.60%, this percentage value shows the closeness of the actual value with the theoretical value. Thus, the percentage value is considered good because the product forming is closed to the maximum amount of the product that could have been produced. However, the reason why the value is less than 100% is because of random error that might happen that is caused by inaccurate measurement scale or spilling of mixture during the experiment.

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CONCLUSION

In conclusion. This experiment for reaction 1 the value obtained for limiting reactant is CaCl2 at 0.01mol and excess reactant is NaCO3 at 0.02mol. As for the percent yield is 88.65%. For reaction 2 the value obtained for limiting reactant is NaCO3 at 0.0075mol and for excess reactant is CaCl2 at 0.01mol. The percent yield is 74.60%. Overall, the yields obtained were satisfied.

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QUESTION & CALCULATION QUESTION 1 Reaction 1

a) Balanced the equation: CaCl2 (aq) + Na2CO3 (aq) → CaCO3 (s) + 2NaCl (aq) b) Determine the limiting reactant

Compound

CaCl2

Na2CO3

Volume (L)

0.02

0.01

No of mole(mol) n=mV

(0.50)(0.02) = 0.01

(1.50)(0.01) = 0.02

mol/coefficient

0.01/1 = 0.01mol

0.02/1 = 0.02mol

Reactant

Limiting Reactant

Excess Reactant

c) Determine the theoretical yield 1mol CaCl2 = 1mol CaCO3 0.01mol CaCl2 = 0.01mol CaCO3

m

= n x mm = 0.01mol x 100.088 = 1.00.1g

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d) % yield

0.8873/1.001 x 100 = 88.65%

REACTION 2

a) Balanced the equation CaCl2 (aq) + naCO3 (aq) → CaCO3 (s) + 2NaCl (aq)

b) Determine the limiting reactant

CaCl2

NaCO3

Volume (L)

0.02

0.005

No of mole(mol) n=mv

0.50 x 0.02 = 0.01

1.50 x 0.005 = 0.0075

mol/coefficient

0.01/1 = 0.01mol

0.0075/1 = 0.00075mol

Reactant

Excess reactant

Limiting reactant

c) Determine the theoretical value

1mol NaCO3 = 1mol CaCO3 0.0075mol NaCO3 = 0.0075 CaCO3

m =n xmm = 0.0075 x 100.088 = 0.750g

d) Percent yield

0.560/0.75066 x 100 = 74.60%

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QUESTION 2 Was the yield you obtained satisfactory? Justify your answer

Yes, based on the percentage value for the reaction 1 and reaction 2 which is 88.65% and 74.60% respectively. These values are considered a high percentage value because the product forming has closed to the maximum amount of product that could have been produced. However, the reason the value is not yet 100% is because of the random error that may happen caused by inaccurate reading measurement scale or spilling of mixture during the experiment.

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REFERENCES

1. Helmenstine, Anne Marie, Ph.D. (2020, October 29). Percent Yield Definition and Formula. Retrieved from https://www.thoughtco.com/definition-of-percent-yield605899 2. Merriam-Webster. (n.d.). Stoichiometry. In Merriam-Webster.com dictionary. Retrieved May 1, 2021, from https://www.merriamwebster.com/dictionary/stoichiometry 3. Anonymous. Limiting reactant and reaction yields (article). https://www.khanacademy.org/science/ap-chemistrybeta/x2eef969c74e0d802:chemicalreactions/x2eef969c74e0d802:stoichiometry/a/limi ting-reagents-and-percent-yield 4. Anonymous. Why is a limiting reactant important in stoichiometry? https://socratic.org/questions/why-is-a-limiting-reactant-important-in-stoichiometry 5. Anonymous.Stoichiometry, Product Yield, and Limiting Reactants https://www.jove.com/science-education/11141/stoichiometry-product-yield-andlimiting-reactants

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