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Stoichiometry of Sodium Hypochlorite & Potassium Iodide: Lab Experiment - Prof. Patricia G, Lab Reports of Chemistry

This document details a laboratory experiment conducted to determine the stoichiometry of sodium hypochlorite and potassium iodide in an aqueous solution. Experimental data, calculations, and results, as well as a discussion of experimental uncertainty.

Typology: Lab Reports

Pre 2010

Uploaded on 11/02/2008

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Download Stoichiometry of Sodium Hypochlorite & Potassium Iodide: Lab Experiment - Prof. Patricia G and more Lab Reports Chemistry in PDF only on Docsity! October 13, 2008 Mat Alburn October 20, 2008 Marcus Sheth Swapnil Stoichiometry Objective: The objective of this lab was to determine the stoichiometry of sodium hypochlorite and potassium iodide in an aqueous solution. Experimental Data: Volume NaClO Volume KI Temp NaClO Temp KI Final Temp 5.0 ml 45ml 22.12˚ 22.20˚ 25.21˚ 10ml 40ml 22.12˚ 22.20˚ 28.90˚ 15ml 35ml 22.12˚ 22.20˚ 27.80˚ 20ml 30ml 22.12˚ 22.20˚ 26.67˚ 25ml 25ml 22.12˚ 22.20˚ 25.40˚ 30ml 20ml 22.12˚ 22.20˚ 25.65˚ 35ml 15ml 22.12˚ 22.20˚ 24.55˚ 40ml 10ml 22.12˚ 22.20˚ 23.58˚ 45ml 5.0ml 22.12˚ 22.20˚ 22.37˚ Sample Calculations: Delta T: Final temp – initial temp final temp=25.21˚ initial temp=22.19˚ 25.21˚ – 22.19˚= 3.02˚ Tinitial: (T ¿¿1V 1)+(T 2V 2) V 1+V 2 ¿ T1=22.12˚ T2=22.20˚ V1=5.0ml V2= 45ml (22.12 ˚∗5 .0ml)+(22.20˚∗45ml) 5.0ml+45ml =22.19 ˚ Mol fraction Ratio: mol fraction NaClO= volume NaClO volume NaClO+volume KI Volume NaClO=5.0ml volume KI=45ml Mol fraction NaClO = 5.0 5.0+45 = 0.10 Point of intersection for the data sets: 1st line = 32x-0.18 2nd line = -8.25x+7.94 32x-0.18 = -8.25x+7.94 X=0.20 Then take x and plug it into either equation to get y 32(0.20)-0.18=y y=6.6 Maximum heat release = (0.20, 6.6) = (mol ratio, ΔT) T) Mole fraction for NaClO= a/(a+b)=0.20 Results and Conclusion: Delta T Mol Fraction Tinitial 3.02˚ 0.10mol 22.19˚ 6.22˚ 0.20 mol 22.18˚ 5.62˚ 0.30 mol 22.18˚ 4.50˚ 0.40 mol 22.17˚ 3.24˚ 0.50 mol 22.16˚ 3.50˚ 0.60 mol 22.15˚ 2.41˚ 0.70 mol 22.14˚ 1.44˚ 0.80 mol 22.14˚ 0.24˚ 0.90 mol 22.13˚ X of the intersection point Y of intersection point X=0.20 Y=6.6 The conclusion I determined from the graph of the two plotted data sets was x=0.20. This was the point on the x axis where the two lines intersected. The x axis stood for mol fraction therefore the mol fraction of the data set is 0.20. in this lab we determined the stoichiometry of an equation by using the hands on approach. Instead of the normal written from we determined the stoichiometry of the equation.
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