Mass Relationships in Chemical Reactions Free Essay Example

Aim

The aim of this experiment is to show that a reaction doesn’t have always 100% yield by reacting NaHCO3 and HCl and determining the amount of the products to calculate actual yield.

Introduction

A chemical reaction will be quantitative if one of the reactants is completely consumed. In this experiment sodium bicarbonate and hydrochloric acid start a reaction. The formula of this reaction is below.

NaHCO3 + HCl –> NaCl + H2O + CO2

Observations

In this experiment, sodium bicarbonate is put in an evaporating dish and some amount of HCl is added in the dish and the reaction started.

Bubbles are formed and CO2 gas is produced and the reaction started to make sound. There was also water vapor formed. White NaHCO3 started to turn into a colorless liquid after adding HCl. As the reaction takes place water is started to form. NaCl was dissolved in water, so salty water is heated to obtain NaCl. As the liquid is heated it turned into a yellowish color for a few seconds.

Then it started bubbling and water vapor is formed.

Raw Data:

Trial #

Mass of Dish+NaHCO3+Lid

+- 0.1 (g)

Mass of NaCl+Water+Dish+Lid

+- 0.1 (g)

Mass of NaCl+Dish+Lid

+- 0.1 (g)

1

64.14 g.

72.16 g.

63.28 g.

2

65.14 g.

72.95 g.

63.91g.

Mass of Evaporating Dish + Lid: 62.14 +-0.1 g

Processed Data:

Trial #1

64.14 – 62.14 = 2 g NaHCO3

72.16 – 62.14 = 10.02 g NaCl + H2O

63.28 – 62.14 = 1.14 g NaCl

Trial # 2

65.14 – 62.14 = 3 g NaHCO3

72.95 – 62.14 = 10.81 g NaCl + H2O

63.91 – 62.14 = 2.07 g NaCl

Trial #

Mass of NaHCO3 (g)

Mass of NaCl + H2O (g)

Mass of NaCl (g)

1

2 g

10.02 g

1.14 g

2

3 g

10.81 g

1.77g

Calculations

Na: 14.01 g/mol, H: 1.01 g/mol, Cl: 35.45 g/mol, O: 16 g/mol, C: 12.01 g/mol

NaCl= 49.46 g/mol

H2O= 18.02 g/mol

NaHCO3: 75.03 g/mol

Mole number of NaHCO3 = mole number of NaCl

Trial #1

2 / 73.03 = 0.0274 mol NaHCO3

1.14 / 49.46 = 0.0230 mol NaCl

Theoretical Yield: 0.0274 mol NaCl

Percent Yield: 0.0230 / 0.0274 = 0.8394 x 100 = 83.94%

Trial #2

3 / 73.03 = 0.0411 mol NaHCO3

1.77 / 49.46 = 0.0358 mol NaCl

Theoretical Yield: 0.0411 mol NaCl

Percent Yield: 0.0358 / 0.0411 = 0.8710 x 100 = 87.10%

Conclusion

The results are 83.94% for trial #1 and 87.10% for trial #2. Trial #2 is more accurate. The accepted value is 100%. The percentage errors are 16.06% for trial #1 and 12.90% for trial #2. The uncertainties are too small to calculate on the results. Random errors presented in this experiment. All the errors were done by human beings. There weren’t any errors due to a flaw of a machine or the procedure.

Evaluation

When salty water is heated on the first trial, the substance started to spill around, because the substance is heated with high amount of heat and faster than it should be. As a result, some of the NaCl which stuck on the lid and spilled around was lost, so the result of the first experiment is not accurate. Other reasons that changed the results may be all NaHCO3 may not be dissolved. Too much HCl may be added on the dish. There may be still water molecules left on the salt after heating. To get more accurate results, the experiment should be done more slowly than this experiment. Especially the heating process should be done slowly, so the evaporation can be observed more carefully.

Remember! This is just a sample.

You can get a custom paper by one of our expert writers.

Get your custom essay

Helping students since 2015