Assessed Practical Titration Write-Up Equation: Na2CO3 + H2SO4 à Na2SO4 + CO2 + H2O One mol of Na2CO3 reacts with one mol of H2SO4. Results: The weight of my sodium carbonate crystals was 2.67g and the results of the titrations are as follows: Rough 1st 2nd 3rd 4th 5th 6th Initial Reading 00.00 00.50 00.00 00.00 00.00 00.00 00.20 Final Reading 26.45 26.45 26.05 27.00 25.85 25.90 26.10 Titration 26.45 25.95 26.05 27.00 25.85 25.90 25.90 pH slightly acidic neutral slightly acidic slightly acidic slightly alkali neutral neutral So the average of the closest three titration results are is: 25.95 + 25.90 + 25.90 / 3 = 25.92 The mass of Na2CO3 I used is 2.67g and the relative molecular mass of Na2CO3 is 106. So the number of mols of Na2CO3 I used was: 2.67 / 106 = 0.0251 mols in 250 cm3 So the concentration of the Na2CO3 solution was: 0.1004 mol dm-3. So in the 25cm3 of Na2CO3 solution I used, there were: 0.0251 / 10 = 0.00251 mols. The equation shows how 1 mol of Na2CO3 reacts with 1 mol of H2SO4, so 0.00251 mols of Na2CO3 reacts with 0.00251 mol of H2SO4. So in the 25.92 cm3 of acid I reacted with the Na2CO3 solution, there are 0.00251 mols. So the concentration of the acid is: ( 0.00251 / 25.92 ) x 1000 = 0.0968 mol dm-3 So the concentration of the acid is 0.0968 mol dm-3. Evaluation: There may have been some limiting factors during the experiment which may have affected the results. The first of these may have been the scales, it is possible that when I was weighing out my Na2CO3
Compress the safety bulb, hold it firmly against the end of the pipette. Then release the bulb and allow it to draw the liquid into the pipette.
taken into account. It is also best to make sure you are working in a
Text Box: CaCO3 + HCl = CaCl2 + CO2 + H2O calcium carbonate + hydrochloric acid = calcium chloride + carbon dioxide + water
Reaction 2: H = 50 x 4.18 x -10.3" H = -2152.7 This value is for 1.37g of calcium oxide, not 56.1g, which is its relative molecular mass. Therefore: H =
Na2S203 (aq) + 2HCl (aq) -> 2NaCl (aq) + H20 (l) + SO2 (g) + S (s)
Determine the reaction order for Na2S2O3 using calculations described in the Background. Show your work. Note that your answer will probably not be an even whole number as it is in the examples.
Albiona Bytyqi Ms. Thorpe English 8 12 February 2016 Introduction A student was running an experiment. She laid out all her materials on a table. The materials she needed to complete this experiment was two containers, table salt, a tablespoon, tap water, and two fresh eggs.
by little to a large excess of water in a bucket. Leave for one hour
2C2H2 (g) + 3O2 (g) 2H2O (g) + 4CO (g) [or] 2C2H2 (g) + O2 (g) 2H2O (g) + 4C (g)
According to the graph on amylase activity at various enzyme concentration (graph 1), the increase of enzyme dilution results in a slower decrease of amylose percentage. Looking at the graph, the amylose percentage decreases at a fast rate with the undiluted enzyme. However, the enzyme dilution with a concentration of 1:3 decreased at a slow rate over time. Additionally, the higher the enzyme dilution, the higher the amylose percentage. For example, in the graph it can be seen that the enzyme dilution with a 1:9 concentration increased over time. However, there is a drastic increase after four minutes, but this is most likely a result of the error that was encountered during the experiment. The undiluted enzyme and the enzyme dilution had a low amylose percentage because there was high enzyme activity. Also, there was an increase in amylose percentage with the enzyme dilution with a 1: 9 concentrations because there was low enzyme activity.
In this, the amount of moles in the sodium hydroxide solution after it has been reacted with the aspirin is found using titration, and then compared with the amount of moles it had without the aspirin being added. The difference in moles is the number of moles of sodium hydroxide that reacted with the aspirin, and therefore the number of moles of
Acid-Base Titration I. Abstract The purpose of the laboratory experiment was to determine equivalence. points, pKa, and pKb points for a strong acid, HCl, titrated with a. strong base, NaOH using a drop by drop approach in order to determine. completely accurate data. The data for this laboratory experiment is as follows.
Na2S2O3 (aq) + 2HCl (aq) → 2NaCl (aq) + S (s) + SO2 (g) + H2O (l)
According to (Khan): A complexometric titration as defined by IUPAC as a volumetric titration where a soluble complex can be formed by titrating a metal ion with a ligand in an aqueous solution and a titrant is one of the reacts used in the titration. [4]
And the symbol equation for it is:. Na2S2O3 + 2HCl, S + SO2 + 2Na + H2O. Before conducting my experiment, I will research into, amongst other things, the factors that affect the rate of a reaction. This is so that I may have enough information to understand the effect of temperature on the rate of a reaction and also gain appropriate understanding to make a suitable prediction as to what the outcome of my experiment will be. Reactions occur when the particles of reactants collide together continuously.