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# Aim: To investigate the reaction between sodium bicarbonate powder and ethanoic acid and discover the concept of limiting reagents.

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Introduction

﻿Introduction Aim: To investigate the reaction between sodium bicarbonate powder and ethanoic acid and discover the concept of limiting reagents. Variables Independent: Mass of sodium bicarbonate (NaHCO3) Dependent: Volume of carbon dioxide gas (CO2) Controlled: Ethanoic Acid (CH3COOH) Chemical Reaction NaHCO3 + CH3COOH = CH3COONa + H2O + CO2 Raw data table Mass of NaHCO3/g Volume of CO2 1st Trial (cm3 ) /+ 1 Volume of CO2 2nd Trial (cm3 ) /+ 1 Volume of CO2 3rd Trial (cm3 ) /+ 1 0.30 83 78 78 0.50 128 132 134 0.70 152 154 152 0.90 154 150 152 1.10 154 153 154 1.30 157 153 153 Qualitative observations As conical flask is swirled, tube containing NaHCO3 falls into the ethanoic solution. Large bubbles appear and the solution produces white foam. After a few minutes, bubbles stop appearing and solution in the conical flask turns clear. ...read more.

Middle

1mol = 0.006285mol + 0.00004 *used the average value out of these groups, uncertainty derived using residuals Grams of CH3COOH = 0.006285 + 0.00004 mol 60.5 = 0.3802 + 0.003g n(CH3COOH) = = =0.006331 + 0.00004 mol of CH3COOH Theoretically, 0.006331 + 0.00004 mol of CH3COOH would react with 0.006331 + 0.00004 mol of NaHCO3 , because the ratio is 1:1. But, this may not be the case in our experiment, therefore we need to find the number of moles of NaHCO3 n(NaHCO3) = = 0.01547 + 0.00005 mol There are only 0.01547 + 0.00005 mol of NaHCO3 but 0.006331 + 0.00004 mol of CH3COOH is required for the 1.3 g of NaHCO3 to react completely. CH3COOH is the limiting reagent We can also say that the number of moles in 10 cm3 of CH3COOH is 0.006331 + 0.00004 mol, as CH3COOH is the limiting reagent and is the one that is completely used up. ...read more.

Conclusion

Therefore, we only need to find the number of moles of NaHCO3 to find the limiting reagent in group 2. n(NaHCO3) = = 0.005952 + 0.00005 mol Theoretically, 0.006331 + 0.00004 mol NaHCO3 is required to completely react with 0.006331 + 0.00004 mol of CH3COOH There are only 0.005952 + 0.00005 mol of NaHCO3 but 0.006331 + 0.00004 mol of CH3COOH NaHCO3 is the limiting reagent in group 2 Now we know that Group 1 NaHCO3 is the limiting reagent Group 2 NaHCO3 is the limiting reagent Group 3 CH3COOH is the limiting reagent Group 4 CH3COOH is the limiting reagent Group 5 CH3COOH is the limiting reagent Group 6 CH3COOH is the limiting reagent We want to know how much sodium bicarbonate in grams is required to react perfectly with 10 cm3 of diluted CH3COOH which contains 0.006331 mol CH3COOH For it to react ideally, 0.006331 mol of NaHCO3 has to react with 0.006331 mol of CH3COOH NaHCO3) = 0.006331 mol 84 = 0.5318 grams ...read more.

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