Analysis of a Chemical Reaction

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Analysis of a Chemical Reaction
Purpose: To observe a chemical reaction and to use qualitative and evidence to identify this reaction among four possibilities. Hypothesis: It is most likely that among the four possibilities that the second one; 2NaHCO3 (s) Na2CO3 (S)+CO2 (g) + H2O (g) will be the correct formula. This is because when acidified, carbonate containing compounds will decompose to form carbon dioxide plus water and the resulting solid product will most likely be Na2CO3. Materials:

* 150 mm rest tubes (2)
* Test tube clamp
* Burner
* Retort stand
* Clay triangle
* Iron ring
* Crucible
* Sodium hydrogen bicarbonate (limewater)

Procedure
Part A
1. All materials were collected
2. 0.5 g of NaHCO3 was added to a test tube.
3. A test tube was obtained which was filled one-quarter with limewater. 4. NaHCO3 was gently heated for 2 minutes over Bunsen burner, while a testube containing lime water was held up to it. 5. Observations were recorded

Part B
1. A data table was prepared
2. Crucible was cleaned and dried.
3. Mass of crucible was determined
4. 1.7g of NaHCO3 was added to the crucible.
5. Crucible was placed on a clay triangle, on a ring stand 6. The substance was heated gently for 1 minute and then heated vigorously for 6 minutes. 7. Crucible cooled for 5 minutes
8. Mass of crucible and residue were determined.

Observations
Part A
* Lime water turned cloudy
* Water droplets at the top of the test tube
Part B
Object/ Substance| Mass (g)|
Mass of crucible| 17.52g|
Mass of crucible and NaHCO3| 19.23g|
Mass of crucible and residue after heating| 18.58g|
Mass of residue| 1.06g|

Analysis / Calculations
1. eq 2) 2NaHCO3 (s)--> Na2CO3 (s)+ CO2 +H20
ratio 2:1

eq 3) 2NaHCO3 (s) --> Na2O (s) + 2CO2 +H2O
ratio 2:1

2. eq 2) 2NaHCO3 (s) Na2CO3 (S)+CO2 (g) + H2O (g)
n= ?
m = 1.7g
MM= 22.99 + 1.01 + 16 (3)
= 84.01 g/mol
n= m / MM
n= 1.7 g / 84.01 g/mol
= 0.020 mol
Eq 3) 2NaHCO3 (s) Na2O (S) +2 CO (g) + H2O (g)
n= ?
m = 1.7g
MM= 22.99 + 1.01 + 16 (3)
= 84.01 g/mol
n= m / MM
n= 1.7g / 84.01 g/mol
= 0.020 mol

3. eq 2) 2NaHCO3 (s) Na2CO3 (S)+CO2 (g) + H2O (g)
2 NaHCO3= 1 Na2CO3
0.020 mol X
= 0.010 mol
m=?
n= 0.010 mol
MM= 23 (2) + 12.01 + 16 (3)
= 106.01 g/mol
m= n x mm
m= 0.010 mol x 106.01 g/mol
= 1.059 g

Eq 3) 2NaHCO3 (s) Na2O (S) +2 CO (g) + H2O (g)

2 NaHCO3= 1 Na2CO3
0.020 mol X
= 0.010 mol
m=?
n= 0.010 mol
MM= 23 (2) + 16
= 62 g/mol
m= n x MM
m= 0.010 mol x 62 g/mol
= 0.62 g
4. Based on the previously collected data and the calculations, it would appear that the second equation 2) 2NaHCO3 (s) Na2CO3 (S)+CO2 (g) + H2O (g) is the correct chemical equation because it has a actual solid product mass of 1.059 g which is only a 0.001g difference of the theoretical yield of 1.06g. As opposed to the third equation 3) 2NaHCO3 (s) Na2O (S) +2 CO (g) + H2O (g) which had a solid product mass of 0.62 g which has a 0.44g difference of the theoretical yield of 1.06g. 5. eq.2) 2NaHCO3 (s) Na2CO3 (S)+CO2 (g) + H2O (g)

%yield =actual/theoretical x 100%
%yield = 1.059/1.06 x 100%
= 99%
%error= (1.06 – 1.059 /1.06) x (100) = 0.094 %

eq. 3) 2NaHCO3 (s) Na2O (S) +2 CO (g) + H2O (g)
%error= actual/theoretical x 100%
% error = 0.062/1.059 x 100%
= 58.4%
%error = ( 1.06 – 0.62 /1.06) x (100) = 42%
Application
The reaction seen in the lab is a chemical decomposition. Decomposition reactions are commonly used in everyday life. For example Air cleaners use decomposition reactions to extract harmful chemicals from the air and replenish them with non-hazardous ones. Another application of decomposition formulas is when attempting to produce CO2; when CaCO3 is heated the resulting product are calcium oxide and carbon dioxide. This exercise is commonly used in today’s chemical industry. The...
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