Post-Lab Report: Mole | MyPaperHub.com

Post-Lab Report: Mole

Post-Lab Report: Mole

Posted on Jun 2018:- By: PaperHub
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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.