Post-Lab Report: Mole

Published on Jun 2018

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

H2C2O_{4 (aq)} +NaOH_(aq) Na_(g) + H₃C₂O_{5 (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 H₃C₂O_5
(g)which is gaseous in nature thus increasing in weight.

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