Post-Lab Report: Mole
Post-Lab Report: Mole
Posted on Jun 2018:- By: PaperHubDisclaimer: This sample paper should be used with proper reference. Email us for citations.
Stoichiometry of a Chemical
Reaction
Post-Lab Report
(a) Report the collected data
with the correct uncertainty.
|
Trial |
Mass H2C2O4 (g) |
Initial
Volume NaOH(mL) |
Final Volume NaOH(mL) |
(NaOH) mol/L |
|
1 |
0.1 |
3 |
20.4 |
0.100 |
|
2 |
0.1996 |
3.5 |
34.6 |
0.100 |
|
3 |
0.1959 |
2.5 |
33.5 |
0.100 |
(b) Calculate the moles of each
reactant used (Show one sample calculation).
Mole
= Mass/Molar mass
Molar Mass of (H2C2O4)
(H=1*2 + C=2*12 + O=4*16) =90
g/mol
Mole
= 0.1g/90 g/mol = 0.00111
|
Trial |
Mol H2C2O4 |
Mol NaOH |
Mol (H2C2O4)/ mol (NaOH) |
|
1 |
0.00111 |
0.03 |
0.037 |
|
2 |
0.00223 |
0.035 |
0.064 |
|
3 |
0.00227 |
0.025 |
0.091 |
(c) Report the balanced chemical
equation.
H2C2O4 (aq) +NaOH(aq) Na(g)
+ H3C2O5 (g)
Report
The
balanced chemical equation is an indication that the both reduction and
addition took place. The sodium hydroxide was reduced into the sodium metal,
thus reducing its mass. However, the crystalline carboxyl acid received two
elements of carbon and four elements of oxygen resulting into a different
compound H3C2O5
(g) which
is gaseous in nature thus increasing in weight.