Stoichiometry of precipitation reactions

STOICHIOMETRY OF PRECIPITATION REACTIONS

University affiliation

Abstract

The concept ofstoichiometry is applied in reactions to determine the amount of thesubstances involved in a reaction. The stoichiometry of precipitationreactions can be used to find out the amount substance thatprecipitates and the amount of substance needed to react with anotherreactant by using mole ratio. In this experiment, the product, CaCO3,which is precipitated out from the reaction between calcium chlorideand sodium carbonate is determine using the moles ratio concept. Thetheoretical weight of the precipitate is found using the moles of thereactants which is then compared with the experimental values.

Experiment andobservation

The procedurewhich I used to conduct this experiment was retrieved from my labpaq.I measured 1.0 g of CaCl2. 2H2O and added it into 25 ml of water in abeaker. The beaker was swirled until it dissolved completely. Usingstoichiometry, the amount of Na2CO3 needed to react 1.0 g of CaCl2.2H2O was calculated. I calculated the theoretical value of sodiumcarbonate needed and found it to be 0.72 g. 0.72 g Na2CO3 was weighedand added to a separate beaker with 25 ml followed by swirling untilit was completely dissolved. Sodium carbonate solution was dispensedinto the beaker with hydrated calcium chloride and swirled until theycompleted mixed. After cloudy color appeared in the solution, it wasfiltered through a filter paper of known weight in order to filterout the CaCO3 precipitate formed. The filter paper was dried andweighed to determine the weight of the precipitate that was recorded.The data collected from the experiment was recorded in the tablebelow.

Object

Weight (g)

Mass of weighing dish

0.6

Mass of weighing dish + sodium carbonate

0.72

Net mass of the sodium carbonate

0.12

Mass of filter paper

1

Mass of filter paper + dry calcium carbonate

1.7

Net mass of the dry calcium carbonate

0.7

Moles of hydrated calcium chloride in 1 g= =0.0068 moles

Na2CO3(aq) + CaCl2. 2H2O(aq)→CaCO3(s) + 2NaCl(aq) + 2 H2O(aq)

CaCl2. 2 H2O(aq) : CaCO3(s) is 1:1hence the moles CaCO3(s) formed is 0.0068 moles

Theoretical yield of CaCO3= moles of CaCO3(s) xrelative formula mass of CaCO3(s)

=0.0068 x 100.06

=0.6804g

CaCO3(s)

Mass (g)

Theoretical

0.6804

Experimental

0.7

Calculation and errors

Since theexperiment involved measurement, it had some errors that affected thedata collected about the precipitate formed. This can be noted by thedifference between the theoretical and experimental values of CaCO3formed after complete reaction between CaCl2.2 H2O and Na2CO3. One ofthe sources of error could have been from inaccurate reading of thegraduated cylinder when measuring distilled water difficulties indetermining whether the level was below or above the meniscus. Theweighing equipment could have also been a source of error as it onlyshowed the values to the nearest hundredth.

Discussion

CaCl2. 2H2O(aq)had Ca2+ and Cl- while Na2CO3(aq) had Na+ and CO32+. When the twosolutions were mixed, there was a reaction that led to the formationof insoluble CaCO3 and NaCl, which was soluble. Na+ and Cl-,spectator ions, were not involved in the precipitation and hence theyleft out of the calculations. The amounts of Ca2+ and CO32+ wereknown hence they result in limiting reagent situation. Theoreticalcalculations were used to find the grams of the Na2CO3 need to reactcompletely CaCl2.2 H2O. Therefore, the reaction between Na2CO3 andCaCl2.2 H2O was assumed to reach completion. The amount ofprecipitate formed from the reaction was above the theoretical value,and this result could be enhanced by the use of the clearly graduatedcylinder and improve the weighing machine.

Questions

Q. A)

=0.68g

Q. B)

=0.7g

Q. C)

The percentyield=(actual yield)/(theoretical yield) X 100

=0.7/0.68×100%

=102 %

Q. D)

Based on myresults, I was fairly accurate. The error could have been that it wasnot completely dried. There is a possibility that the temperature andthe humidity may have been the error.

Q. E)

These errorscould be decreased in the future by leaving it out longer to makesure that it is completely dry and, also, try to store it where thetemperature is constant.

Q. F)

a)

Moles of CaCl2. 2H2O in 1.0 g is 0.0068 moles

Moles of Na2CO3in 1.0 g is 1/105.988= 0.0094 moles

The amount ofCaCO3 produced, in this case, will be still the same, 0.68 g. Themole ratio for the reaction between the two reagents is 1:1 and hencemoles of the two have to be the same for both of them to becompletely consumed. The reaction stopped when all the 0.0068 molesof CaCl2. 2 H2O were consumed, and no more CaCO3 can be formed.

b)

CaCO3 was thelimiting reagent while Na2CO3 was in excess. Since the mole ratio ofCaCO3 and Na2CO3 is 1:1, the moles of Na2CO3 that reacted were 0.0068moles. Excess moles are 0.0094-0.0068= 0.0026 moles.

Excess mass=105.988 x 0.0026= 0.27g

Q. G)

Mass of acetic anhydride= 3.2 x 1.08= 3.456g

Moles of aceticanhydride= 3.456/102=0.03388 moles

Moles ofsalicylic acid= 1.45/138=0.01050 moles

The limitingreactant was salicylic acid.

Theoretical yieldis 0.0105 moles x 180 (molecular weight) = 1.89 g

Percentageyield=1.23/1.89=65.07 %

References

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ZUMDAHL, S. T. E. V. E. N. S., &amp ZuMDAHL, S. A.(2009).&nbspChemistry: 7th Edition. Boston: Houghton Miffin.

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