# Stoichiometry Reaction

Pages: 2 (285 words) Published: November 5, 2010
Title:
Stoichiometry Reaction
Objectives:
1.To decompose sodium hydrogen carbonate (sodium bicarbonate) by heating. 2.To accurately measure the degree of completion of the reaction by analysing the solid sodium carbonate product. 3.To calculate amount of product with given amount of reactant. 4.To determine amount of heat release in the reaction.

Results:
Part 1: Thermal Decomposition of NaHCO3
MaterialsMass (g)
Clean and dry test tube15.1632
Clean test tube + NaHCO317.1647
Amount of NaHCO3 added 2.0015
After heated test tube + NaHCO316.4500
Amount of solid product (NaHCO3 + any unreacted NaHCO3) 1.2868

Part 2: Titration of Na2CO3 with Hydrochloric acid (HCl)
Initial volume (cm3)5.0 8.1 22.5
Final volume (cm3)27.7 30.7 45.2
Volume of HCL used (cm3)22.7 22.6 22.7

Calculations:
Part 1
2NaHCO3 (s) ——> Na2CO3(s) + CO2 (g) + H2O (l)
From equation, 2 mol of NaHCO3 will produce 1 mol of Na2CO3 Mol of NaHCO3= 2.0015g / 84gmol-1
=0.0238 mol
=0.024 mol
Mol of Na2CO3 = 0.024 mol x ½
= 0.012 mol
Experimental mass of Na2CO3= 0.012 mol x 106 gmol-1
= 1.272g (experimental mass)
Theory mass =1.2868g
Percentage of yield (%) = (Experimental mass / Theory mass ) x 100%
= (1.272g/2.2868g) x 100%
= 98.85 %

Part 2
Na2CO3 (aq) + HCl (aq) ——> NaHCO3 (aq) + NaCl (aq)

M1V1 1
-------------- = ------
M2V2 1
(0.05)(22.67)
--------------- = M2 = 0.04534 M
M2 (25.00)

MV (0.04534M) (250 ml)
No. of mol = -------- = -----------------------------
10001000
= 0.011335 mol

Mass= Molar mass x Number of moles
=0.011335 x 106
1.2015 g
= -------------- x 100% = 93.37%
1.2868 g

Please join StudyMode to read the full document