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# Limiting Reagents. 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. ...read more.

Middle

cm3 3 = 79.66667 cm3 Rounded to two significant figures is 80 cm3 Calculating the uncertainty for my averages Calculated using residuals i.e. Largest residual is 83 - 80 = 3 as opposed to 80 - 78 = 2 Absolute uncertainty is + 3 The limiting reagent as when CO2 remains roughly the same (group 3 to 6) Because Molar Volume of a gas = 24.5 dm3 at SLC (approximately) NaHCO3 + CH3COOH = CH3COONa + H2O + CO2 1 mole of CO2 is 24500 cm3 and (153 cm3 + 1.0 cm3)* of CO2 is 0.006244 moles because (153 + 0.6% 24500) 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) ...read more.

Conclusion

However, it is noteworthy that the task sheet said that it is a 3-5% acid solution, therefore the uncertainty of the number of moles of CH3COOH could be larger than expected. Group 1 We know that the number of moles in 10 cm3 of CH3COOH is 0.006331 + 0.00004 mol as seen before. Therefore, we only need to find the number of moles of NaHCO3 to find the limiting reagent in group 1. n(NaHCO3) = = 0.003571 + 0.00005 mol Theoretically, 0.006331 + 0.00004 mol NaHCO3 is required to completely react with 0.006331 + 0.00004 mol of CH3COOH But there are only 0.003571 + 0.00005 mol of NaHCO3 but 0.006331 + 0.00004 mol of CH3COOH NaHCO3 is the limiting reagent in group 2 ...read more.

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