• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

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

Extracts from this document...

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.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our International Baccalaureate Chemistry section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Chemistry essays

  1. Chemistry Limiting Reactant Lab

    Furthermore, because there are uncertainties in all the equipment used, there is a margin of error associated with each apparatus. Nevertheless, the margin of error increases when performing various calculations such as concentration and volume. Therefore all calculations will not be perfectly accurate resulting in discrepancies in comparison to theory.

  2. Aim: Using an iodine clock reaction to find the order of hydrogen peroxide and ...

    An example of this is while changing the volume of H2O2; the other reactants' volumes should be held constant, and H2O2's volume should increase by 2 for each experiment. 10. Carry out the experiment with volumes of 2, 4, 6, 8 and 10 for both H2O2 and CH2COOH while keeping

  1. Limiting Reagents Lab

    precipitate = .69g Theoretical mass of Cu (s) precipitate = .719g % yield = 96% Conclusion and Evaluation (CE) Aspect 1: Conclusion: The mass of precipitate that forms when 1.93g of CuCl2.2H20 reacts with 1.31g of aluminum is .69 g.

  2. Aim To investigate the chemical properties of alcohol

    10 drops of dilute sulphuric acid + 5 drops of potassium dichromate + 5 drops of ethanol Initially a colourless solution is present. Upon heating the solution turns gradually to a dark green colour. Discussion In the reaction between sodium and ethanol the products are sodium ethoxide and hydrogen gas.

  1. Percent Yield Lab. This experiment has proven that KI is the limiting reagent ...

    The theoretical mole is 0.00602 moles and the theoretical mass is 2078g. Equipment Potassium Iodide(2.00g) Lead Nitrate (3.00g) Distilled water (200ml) Beakers (300ml *5) Goggles Glass Stirring Rod Filter Paper (small x2 large x1) Funnel Retort Stand Ring Clamp Scale Procedure: 1.

  2. Aim: To investigate the reaction between sodium bicarbonate powder and ethanoic acid and discover ...

    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)

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work